Example 1

Problem : A projectile is thrown with a speed "u" at an angle 60° over an incline of 30°. If the time of flight of the projectile is “T”, then find the range of the flight.

Figure 1: The time of flight is “T”.

Projectile motion on an incline

Solution : We can see here that time of flight is already given. We can find range considering projectile motion in the coordinates of horizontal and vertical axes. The range of the projectile “R” is obtained by using trigonometric ratio in triangle OAB. The range is related to horizontal base “OB” as :

Figure 2: The time of flight is “T”.

Projectile motion on an incline

cos 30 0 = O B O A cos 30 0 = O B O A

R = O A = O B cos 30 0 = O B sec 30 0 R = O A = O B cos 30 0 = O B sec 30 0

Now, we can find OB by considering motion in horizontal direction :

⇒ R = O B = u x T = u cos 60 0 T = u T 2 ⇒ R = O B = u x T = u cos 60 0 T = u T 2

Thus, the range of the projectile, OA, is :

⇒ R = O A = u T sec 30 0 2 = u T 3 ⇒ R = O A = u T sec 30 0 2 = u T 3

Example 2

Problem : A particle is projected from the foot of an incline of angle “30°” at a certain velocity so that it strikes the incline normally. Find the angle of projection (θ ) as measured from the horizontal.

Figure 3: The projectile hits the incline normally.

Projectile motion on an incline

Solution : Here, projectile hits the incline normally. It means that component of velocity along the incline is zero. We should remember that the motion along the incline is not uniform motion, but a decelerated motion. In order to take advantage of the fact that final component velocity along incline is zero, we consider motion in a coordinate system along the incline and along a direction perpendicular to it.

Figure 4: The projectile hits the incline normally.

Projectile motion on an incline

v x = u x + a x T v x = u x + a x T

⇒ 0 = u cos θ − 30 0 − g sin 30 0 T ⇒ 0 = u cos θ − 30 0 − g sin 30 0 T

T = u cos θ − 30 0 g sin 30 0 T = u cos θ − 30 0 g sin 30 0

Now, time of flight is also given by the formulae :

⇒ T = 2 u sin θ − 30 0 g cos 30 0 ⇒ T = 2 u sin θ − 30 0 g cos 30 0

Equating two expressions for time of flight, we have :

⇒ 2 u sin θ − 30 0 g cos 30 0 = u cos θ − 30 0 g sin 30 0 ⇒ 2 u sin θ − 30 0 g cos 30 0 = u cos θ − 30 0 g sin 30 0

⇒ tan θ − 30 0 = 1 2 tan 30 0 = 3 2 ⇒ tan θ − 30 0 = 1 2 tan 30 0 = 3 2

⇒ θ = 30 0 + tan - 1 3 2 ⇒ θ = 30 0 + tan - 1 3 2

Example 3

Problem : A ball is projected on an incline of 30° from its base with a speed 20 m/s, making an angle 60° from the horizontal. Find the speed with which the ball hits the incline.

Figure 5: The projectile is projected from base of the incline.

Projectile motion on an incline

Solution : We analyze this problem in the coordinates along the incline (x-axis) and in the direction perpendicular to the incline (y-axis). In order to find the speed at the end of flight, we need to find the component velocities in “x” and “y” directions.

The velocity in y-direction can be determined making use of the fact that a ball under constant acceleration like gravity returns to the ground with the same speed, but in opposite direction. The component of velocity in y-direction at the end of the journey, therefore, is :

Figure 6: The projectile is projected from base of the incline.

Projectile motion on an incline

v y = − u y = − 20 sin 30 0 = − 20 X 1 2 = - 10 m / s v y = − u y = − 20 sin 30 0 = − 20 X 1 2 = - 10 m / s

Now, we should attempt to find the component of velocity in x-direction. We should, however, recall that motion in x-direction is not a uniform motion, but has deceleration of “-gsinα”. Using equation of motion in x-direction, we have :

v x = u x + a x T v x = u x + a x T

Putting values,

v x = u cos 30 0 − g sin 30 0 T = 20 X 3 2 − 10 X 1 2 X T = 10 3 − 5 T v x = u cos 30 0 − g sin 30 0 T = 20 X 3 2 − 10 X 1 2 X T = 10 3 − 5 T

Clearly, we need to know time of flight to know the component of velocity in x-direction. The time of flight is given by :

T = 2 u y a y = 2 u sin 30 0 g cos 30 0 = 2 X 20 3 X 10 = 4 3 T = 2 u y a y = 2 u sin 30 0 g cos 30 0 = 2 X 20 3 X 10 = 4 3

Hence, component of velocity in x-direction is :

⇒ v x = 10 3 − 5 T = 10 3 − 5 X 4 3 ⇒ v x = 10 3 − 5 T = 10 3 − 5 X 4 3

v x = 30 - 20 3 = 10 3 v x = 30 - 20 3 = 10 3

The speed of the projectile is equal to resultant of components :

⇒ v = { - 10 2 + 10 3 2 } ⇒ v = { - 10 2 + 10 3 2 }

v = 400 3 = 20 3 m / s v = 400 3 = 20 3 m / s

Example 4

Problem : A ball falls through a height “H” and impacts an incline elastically. Find the time of flight between first and second impact of the projectile on the incline.

Solution : The ball falls vertically through a distance "H". We can get the value of initial speed by considering free fall of the ball before impacting incline.

⇒ 0 = u 2 − 2 g H ⇒ 0 = u 2 − 2 g H

⇒ u = 2 g H ⇒ u = 2 g H

In order to answer this question, we need to identify the velocity with which projectile rebounds. Since impact is considered elastic, the projectile is rebounded without any loss of speed. The projectile is rebounded such that angle of incidence i.e. the angle with the normal is equal to angle of reflection.

Figure 7: The ball impacts incline elastically.

Projectile motion on an incline

The components of velocity along the incline and perpendicular to it are shown in the figure. The motion of ball, thereafter, is same as that of a projectile over an incline. Here, we shall analyze motion in y-direction (normal to the incline) to find the time of flight. We note that the net displacement between two strikes is zero in y – direction.

Applying equation of motion

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

⇒ 0 = u cos α T − 1 2 g cos α T 2 ⇒ 0 = u cos α T − 1 2 g cos α T 2

T = 0 T = 0

Or

T = 2 u cos α g cos α = 2 u g T = 2 u cos α g cos α = 2 u g

Putting value of initial speed in the equation of time of flight, we have :

⇒ T = 8 H g ⇒ T = 8 H g

It should be noted here that we can find the time of flight also by using standard formula of time of flight for projectile motion down the incline. The time of flight for projection down an incline is given as :

T = 2 u sin θ + α g cos α T = 2 u sin θ + α g cos α

We need to be careful while appropriating angles in the above expression. It may be recalled that all angles are measured from the horizontal. We redraw the figure to denote the value of angle of projection “θ” from the horizon.

Figure 8: The ball impacts incline elastically.

Projectile motion on an incline

⇒ T = 2 u sin 90 0 - 2 α + α g cos α = 2 u sin 90 0 - α g cos α ⇒ T = 2 u sin 90 0 - 2 α + α g cos α = 2 u sin 90 0 - α g cos α

⇒ T = 2 u cos α g cos α = 2 u g = 8 H g ⇒ T = 2 u cos α g cos α = 2 u g = 8 H g

Example 5

Problem : Two incline plane of angles 30° and 60° are placed touching each other at the base as shown in the figure. A projectile is projected at right angle with a speed of 10√3 m/s from point “P” and hits the other incline at point “Q” normally. Find the linear distance between PQ.

Figure 9: A projectile projected at right angle from an incline hits another incline at right angle.

Projectile motion on two inclines

Solution : We notice here OPQ forms a right angle triangle at “O”. The linear distance, “PQ” is related as :

Figure 10: A projectile projected at right angle from an incline hits another incline at right angle.

Projectile motion on two inclines

P Q 2 = O P 2 + O Q 2 P Q 2 = O P 2 + O Q 2

In order to find “PQ”, we need to know “OP” and “OQ”. We can find “OP”, considering motion in y-direction.

y = O P = u y T + 1 2 a y T 2 = 0 + 1 2 a y T 2 = 1 2 a y T 2 y = O P = u y T + 1 2 a y T 2 = 0 + 1 2 a y T 2 = 1 2 a y T 2

Similarly,

x = O Q = u x T + 1 2 a x T 2 = 10 3 T + 1 2 a x T 2 x = O Q = u x T + 1 2 a x T 2 = 10 3 T + 1 2 a x T 2

In order to evaluate these two relations, we need to find components of accelerations and time of flight.

Considering first incline, we have :

a x = - g cos 30 0 = - 10 X 3 2 = - 5 3 m / s 2 a x = - g cos 30 0 = - 10 X 3 2 = - 5 3 m / s 2

a y = - g sin 30 0 = - 10 X 1 2 = - 5 m / s 2 a y = - g sin 30 0 = - 10 X 1 2 = - 5 m / s 2

In order to find the time of flight, we can further use the fact that the component of velocity in x-direction i.e. along the second incline is zero. This, in turn, suggests that we can analyze motion in x-direction to obtain time of flight.

In x-direction,

v x = u x + a x T v x = u x + a x T

⇒ 0 = u x + a x T ⇒ 0 = u x + a x T

⇒ T = - u x a x ⇒ T = - u x a x

Putting values in the equation and solving, we have :

⇒ T = - 10 3 − 5 3 = 2 s ⇒ T = - 10 3 − 5 3 = 2 s

Now, we can evaluate “x” and “y” displacements as :

⇒ y = O P = 1 2 a y T 2 = - 1 2 X 5 X 2 2 = - 10 m ⇒ y = O P = 1 2 a y T 2 = - 1 2 X 5 X 2 2 = - 10 m

⇒ x = O Q = 10 3 x 2 + 1 2 - 5 3 X 2 2 ⇒ x = O Q = 10 3 x 2 + 1 2 - 5 3 X 2 2

x = O Q = 10 3 x = O Q = 10 3

Considering positive values, the linear distance, “PQ” is given as :

P Q = O P 2 + O Q 2 = { 10 2 + 10 3 2 } = 20 m P Q = O P 2 + O Q 2 = { 10 2 + 10 3 2 } = 20 m