Example 1

Problem : The speed of a particle, projected at 60°, is 20 m/s at the time of projection. Find the time interval for projectile to loose half its initial speed.

Solution : Here, we see that final and initial speeds (not velocity) are subject to given condition. We need to use the given condition with appropriate expressions of speeds for two instants.

u = u x 2 + u y 2 u = u x 2 + u y 2

v = v x 2 + v y 2 v = v x 2 + v y 2

According to question,

u = 2 v u = 2 v

⇒ u 2 = 4 v 2 ⇒ u 2 = 4 v 2

⇒ u x 2 + u y 2 = 4 v x 2 + v y 2 ⇒ u x 2 + u y 2 = 4 v x 2 + v y 2

But, horizontal component of velocity remains same. Hence,

⇒ u x 2 + u y 2 = 4 u x 2 + v y 2 ⇒ u x 2 + u y 2 = 4 u x 2 + v y 2

Rearranging for vertical velocity :

v y 2 = u y 2 - 3 u x 2 = 20 sin 60 0 2 - 3 X 20 cos 60 0 2 v y 2 = u y 2 - 3 u x 2 = 20 sin 60 0 2 - 3 X 20 cos 60 0 2

v y 2 = 20 X 3 4 - 3 X 20 X 1 4 = 0 v y 2 = 20 X 3 4 - 3 X 20 X 1 4 = 0

The component of velocity in vertical direction becomes zero for the given condition. This means that the projectile has actually reached the maximum height for the given condition. The time to reach maximum height is half of the time of flight :

t = u sin θ g = 20 X sin 60 0 10 = 3 s t = u sin θ g = 20 X sin 60 0 10 = 3 s

Example 2

Problem : A man can throw a ball to a greatest height denoted by "h". Find the greatest horizontal distance that he can throw the ball (consider g = 10 m / s 2 m / s 2 ).

Solution : The first part of the question provides the information about the initial speed. We know that projectile achieves greatest height in vertical throw. Let "u" be the initial speed. We can, now, apply equation of motion " v 2 = u 2 + 2 g y v 2 = u 2 + 2 g y " for vertical throw. We use this form of equation as we want to relate initial speed with the greatest height.

Here, v = 0; a = -g

⇒ 0 = u 2 - 2 g h u 2 = 2 g h ⇒ 0 = u 2 - 2 g h u 2 = 2 g h

The projectile, on the other hand, attains greatest horizontal distance for the angle of projection, θ = 45°. Accordingly, the greatest horizontal distance is :

R max = u 2 sin 2 X 45° g = u 2 sin 90° g = u 2 g = 2 g h g = 2 h R max = u 2 sin 2 X 45° g = u 2 sin 90° g = u 2 g = 2 g h g = 2 h

Example 3

Problem : A bullet from a gun is fired at a muzzle speed of 50 m/s to hit a target 125 m away at the same horizontal level. At what angle from horizontal should the gun be aimed to hit the target (consider g = 10 m / s 2 m / s 2 ) ?

Solution : Here, horizontal range is given. We can find out the angle of projection from horizontal direction, using expression of horizontal range :

R = u 2 sin 2 θ g ⇒ sin 2 θ = g R u 2 = 10 x 125 50 2 = 1 2 ⇒ sin 2 θ = sin 30 0 ⇒ θ = 15 0 R = u 2 sin 2 θ g ⇒ sin 2 θ = g R u 2 = 10 x 125 50 2 = 1 2 ⇒ sin 2 θ = sin 30 0 ⇒ θ = 15 0

Example 4

Problem : A projectile has same horizontal range for a given projection speed for the angles of projections θ 1 θ 1 and θ 2 ( θ 2 > θ 1 ) θ 2 ( θ 2 > θ 1 ) with the horizontal. Find the ratio of the times of flight for the two projections.

Solution : The ratio of time of flight is :

T 1 T 2 = u sin θ 1 u sin θ 2 = sin θ 1 sin θ 2 T 1 T 2 = u sin θ 1 u sin θ 2 = sin θ 1 sin θ 2

For same horizontal range, we know that :

θ 2 = ( 90 0 - θ 1 ) θ 2 = ( 90 0 - θ 1 )

Putting this, we have :

T 1 T 2 = sin θ 1 sin ( 90 0 - θ 1 ) = tan θ 1 T 1 T 2 = sin θ 1 sin ( 90 0 - θ 1 ) = tan θ 1

Example 5

Problem : A projectile, thrown at an angle 15° with the horizontal, covers a horizontal distance of 1000 m. Find the maximum distance the projectile can cover with the same speed (consider g = 10 m / s 2 m / s 2 ).

Solution : The range of the projectile is given by :

R = u 2 sin 2 θ g R = u 2 sin 2 θ g

Here, R = 1000 m, g = 10 m / s 2 m / s 2 , θ = 15°

⇒ 1000 = u 2 sin ( 2 x 15 0 ) g = u 2 2 g ⇒ u 2 2 g = 1000 m ⇒ 1000 = u 2 sin ( 2 x 15 0 ) g = u 2 2 g ⇒ u 2 2 g = 1000 m

Now, the maximum range (for θ = 45°) is :

⇒ R max = u 2 g = 2000 m ⇒ R max = u 2 g = 2000 m

Example 6

Problem : Two balls are projected from the same point in the direction inclined at 60° and 30° respectively with the horizontal. If they attain the same height, then the ratio of speeds of projection is :

Solution : Since the projectiles attain same height,

Figure 1

Projectile motion

H 1 = H 2 = H u 1 2 sin 2 60 0 2 g = u 2 2 sin 2 30 0 2 g u 1 2 X 3 4 = u 2 2 X 1 4 u 1 2 : u 2 2 = 1 : 3 ⇒ u 1 : u 2 = 1 : 3 H 1 = H 2 = H u 1 2 sin 2 60 0 2 g = u 2 2 sin 2 30 0 2 g u 1 2 X 3 4 = u 2 2 X 1 4 u 1 2 : u 2 2 = 1 : 3 ⇒ u 1 : u 2 = 1 : 3

Example 7

Problem : A projectile is thrown vertically up, whereas another projectile is thrown at an angle θ with the vertical. Both of the projectiles stay in the air for the same time (neglect air resistance). Find the ratio of maximum heights attained by two projectiles.

Solution : Let u 1 u 1 and u 2 u 2 be the speeds of projectiles for vertical and non-vertical projections. The times of the flight for vertical projectile is given by :

T 1 = 2 u 1 g T 1 = 2 u 1 g

We note here that the angle is given with respect to vertical - not with respect to horizontal as the usual case. As such, the expression of time of flight consists of cosine term :

T 2 = 2 u 2 cos θ g T 2 = 2 u 2 cos θ g

As, time of flight is same,

2 u 1 g = 2 u 2 cos θ g u 1 = u 2 cos θ 2 u 1 g = 2 u 2 cos θ g u 1 = u 2 cos θ

On the other hand, the maximum heights attained in the two cases are :

H 1 = u 1 2 2 g H 2 = u 2 2 cos 2 θ 2 g H 1 = u 1 2 2 g H 2 = u 2 2 cos 2 θ 2 g

Using the relation u 1 = u 2 cos θ u 1 = u 2 cos θ as obtained earlier, we have :

H 2 = u 1 2 2 g ⇒ H 1 = H 2 H 2 = u 1 2 2 g ⇒ H 1 = H 2

Example 8

Problem : The times for attaining a particular vertical elevation during projectile motion are t 1 t 1 and t 2 t 2 . Find time of flight, T, in terms of t 1 t 1 and t 2 t 2 .

Solution : We can answer this question analytically without using formula. Let the positions of the projectile at two time instants be "A" and "B", as shown in the figure. The time periods t 1 t 1 and t 2 t 2 denotes time taken by the projectile to reach points "A" and "B" respectively. Clearly, time of flight, T, is equal to time taken to travel the curve OAB ( t 2 t 2 ) plus the time taken to travel the curve BC.

Figure 2

Projectile motion

Now, projectile takes as much time to travel the curve OA, as it takes to travel curve BC. This is so, because the time of travel of equal vertical displacement in either direction (up or down) in vertical motion under gravity is same. Since the time of travel for curve OA is t 1 t 1 , the time of travel for curve BC is also t 1 t 1 . Thus, the total time of flight is :

T = t 1 + t 2 T = t 1 + t 2

Alternatively,

As the heights attained are equal,

h 1 = h 2 ⇒ u y t 1 - 1 2 g t 1 2 = u y t 2 - 1 2 g t 2 2 ⇒ u y ( t 2 - t 1 ) = 1 2 g ( t 2 2 - t 1 2 ) = 1 2 g ( t 2 + t 1 ) ( t 2 - t 1 ) ⇒ t 2 + t 1 = 2 u y g = T h 1 = h 2 ⇒ u y t 1 - 1 2 g t 1 2 = u y t 2 - 1 2 g t 2 2 ⇒ u y ( t 2 - t 1 ) = 1 2 g ( t 2 2 - t 1 2 ) = 1 2 g ( t 2 + t 1 ) ( t 2 - t 1 ) ⇒ t 2 + t 1 = 2 u y g = T

Example 9

Problem : The position of a projectile projected from the ground is :

x = 3 t y = ( 4 t - 2 t 2 ) x = 3 t y = ( 4 t - 2 t 2 )

where “x” and “y” are in meters and “t” in seconds. The position of the projectile is (0,0) at the time of projection. Find the speed with which the projectile hits the ground.

Solution : When the projectile hits the ground, y = 0,

0 = ( 4 t - 2 t 2 ) ⇒ 2 t 2 - 4 t = t ( 2 t - 2 ) = 0 ⇒ t = 0 , t = 2 s 0 = ( 4 t - 2 t 2 ) ⇒ 2 t 2 - 4 t = t ( 2 t - 2 ) = 0 ⇒ t = 0 , t = 2 s

Here t = 0 corresponds to initial condition. Thus, projectile hits the ground in 2 s. Now velocities in two directions are obtained by differentiating given functions of the coordinates,

v x = đ x đ t = 3 v y = đ y đ t = 4 - 4 t v x = đ x đ t = 3 v y = đ y đ t = 4 - 4 t

Now, the velocities for t = 2 s,

v x = đ x đ t = 3 m / s v y = đ y đ t = 4 - 4 x 2 = - 4 m / s v x = đ x đ t = 3 m / s v y = đ y đ t = 4 - 4 x 2 = - 4 m / s

The resultant velocity of the projectile,

v = ( v x 2 + v y 2 ) = { 3 2 + ( - 4 ) 2 } = 5 m / s v = ( v x 2 + v y 2 ) = { 3 2 + ( - 4 ) 2 } = 5 m / s

Example 10

Problem : A projectile, thrown from the foot of a triangle, lands at the edge of its base on the other side of the triangle. The projectile just grazes the vertex as shown in the figure. Prove that :

Figure 3: The projectile grazes the vertex of the triangle.

Projectile motion

tan α + tan β = tan θ tan α + tan β = tan θ

where “θ” is the angle of projection as measured from the horizontal.

Solution : In order to expand trigonometric ratio on the left side, we drop a perpendicular from the vertex of the triangle “A” to the base line OB to meet at a point C. Let x,y be the coordinate of vertex “A”, then,

Figure 4: The projectile grazes the vertex of the triangle.

Projectile motion

tan α = A C O C = y x tan α = A C O C = y x

and

tan β = A C B C = y R - x tan β = A C B C = y R - x

Thus,

tan α + tan β = y x + y R - x = y R x R - x tan α + tan β = y x + y R - x = y R x R - x

Intuitively, we know the expression is similar to the expression involved in the equation of projectile motion that contains range of projectile,

⇒ y = x tan θ 1 - x R ⇒ y = x tan θ 1 - x R

⇒ tan θ = y R x R - x ⇒ tan θ = y R x R - x

Comparing equations,

⇒ tan α + tan β = tan θ ⇒ tan α + tan β = tan θ

Example 11

Problem : A projectile is projected at angle “θ” from the horizontal at the speed “u”. If an acceleration of “g/2” is applied to the projectile due to wind in horizontal direction, then find the new time of flight, maximum height and horizontal range.

Solution : The acceleration due to wind affects only the motion in horizontal direction. It would, therefore, not affect attributes like time of flight or maximum height that results exclusively from the consideration of motion in vertical direction. The generic expressions of time of flight, maximum height and horizontal range of flight with acceleration are given as under :

T = 2 u y g T = 2 u y g

H = u y 2 2 g = g T 2 4 H = u y 2 2 g = g T 2 4

R = u x u y g R = u x u y g

The expressions above revalidate the assumption made in the beginning. We can see that it is only the horizontal range that depends on the component of motion in horizontal direction. Now, considering accelerated motion in horizontal direction, we have :

x = R ′ = u x T + 1 2 a x T 2 x = R ′ = u x T + 1 2 a x T 2

⇒ R ′ = u x T + 1 2 g 2 T 2 ⇒ R ′ = u x T + 1 2 g 2 T 2

R ′ = R + H R ′ = R + H

Example 12

Problem : A projectile is projected at angle “θ” from the horizontal at the speed “u”. If an acceleration of g/2 is applied to the projectile in horizontal direction and a deceleration of g/2 in vertical direction, then find the new time of flight, maximum height and horizontal range.

Solution : The acceleration due to wind affects only the motion in horizontal direction. It would, therefore, not affect attributes resulting exclusively from the consideration in vertical direction. It is only the horizontal range that will be affected due to acceleration in horizontal direction. On the other hand, deceleration in vertical direction will affect all three attributes.

1: Time of flight

Let us work out the effect on each of the attribute. Considering motion in vertical direction, we have :

y = u y T + 1 2 a y T 2 y = u y T + 1 2 a y T 2

For the complete flight, y = 0 and t = T. Also,

a y = - g + g 2 = - 3 g 2 a y = - g + g 2 = - 3 g 2

Putting in the equation,

⇒ 0 = u y T - 1 2 X 3 g 2 X T 2 ⇒ 0 = u y T - 1 2 X 3 g 2 X T 2

Neglecting T = 0,

⇒ T = 4 u y 3 g = 4 u sin θ 3 g ⇒ T = 4 u y 3 g = 4 u sin θ 3 g

2: Maximum height

For maximum height, v y = 0 v y = 0 ,

0 = u y 2 − 2 X 3 g 2 X H 0 = u y 2 − 2 X 3 g 2 X H

H = u y 2 3 g = u 2 sin 2 θ 3 g H = u y 2 3 g = u 2 sin 2 θ 3 g

2: Horizontal range

Now, considering accelerated motion in horizontal direction, we have :

x = R = u x T + 1 2 a x T 2 x = R = u x T + 1 2 a x T 2

R = u x 4 u y g + 1 2 g 2 4 u y g 2 R = u x 4 u y g + 1 2 g 2 4 u y g 2

⇒ R = 4 u y g [ u x + 1 2 g 2 4 u y g ] ⇒ R = 4 u y g [ u x + 1 2 g 2 4 u y g ]

⇒ R = 4 u y g { u x + u y } ⇒ R = 4 u y g { u x + u y }

⇒ R = 4 u 2 sin θ g [ cos θ + sin θ ] ⇒ R = 4 u 2 sin θ g [ cos θ + sin θ ]