A tennis player wins a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/s and at an angle 45° above the horizontal . On its way down, the ball is caught by a spectator 10 m above the point where the ball was hit. (a) Calculate the time it takes the tennis ball to reach the spectator. (b) What are the magnitude and direction of the ball’s velocity at impact?
Answers
Answer:
a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/s and at an angle 45° above the horizontal . On its way down, the ball is caught by a spectator 10 m above the point where the ball was hit.
Given: velocity = 30 m/s
Angle with horizontal = 45°
Range = 10m
To find: (a) Time taken by the ball to reach the spectator
(b) Magnitude and direction of ball's velocity at impact
Solution:
(a) In projectile motion, when an object is thrown with velocity u, and at an angle θ, then the time taken by the object to carry its total path
The time taken can be calculated by = 2uSin θ /g
Putting the values in the equation, time of flight = 2×30×Sin45/10
=3√2 seconds.
Therefore, the time it takes the tennis ball to reach the spectator is 3√2 seconds.
(b) Velocity at impact is the velocity with which the ball lands.
It can be calculated by the equation of kinetics. v² = u² + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration, s is the distance.
u = 30Cos45°
= 30/√2 = 15√2 m/s
Horizontal velocity remains constant.
v = (15√2)² + 2×10×10
= 650 m/s
Therefore, the ball's velocity at impact is 650 m/s.