The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²
The direction of the acceleration of the ball is downwards
The given parameters
initial velocity of the ball, u = 0
height above the ground, h = 2.2 m
time of motion of the ball, t = 96 ms = 0.096 s
The magnitude of the acceleration of the ball while coming to rest is calculated as;
let the downwards direction of the acceleration be positive
The direction of the acceleration of the ball is downwards
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Answer:
1) Position time graph
2) Acceleration time graph
3) Velocity time graph
The second option is the correct one. m/s^2
(a)
consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.
m = mass of the tennis ball = 60 g = 0.060 kg
v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s
v = final velocity of the tennis ball after being hit by racket = - 39 m/s
ΔP = change in momentum of the ball
change in momentum of the ball is given as
ΔP = m (v - v₀)
inserting the above values
ΔP = (0.060) (- 39 - 20)
ΔP = - 3.54 kgm/s
hence , magnitude of change in momentum : 3.54 kgm/s
