Atennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. (a) what is the acceleration of the ball while it is in flight? (b) what is the velocity of the ball when it reaches its maximum height? find (c) the initial velocity of the ball and (d) the maximum height it reaches.

a) - 9.8 m/s²

b) 0 m/s

c) 9.8 m/s

d) 4.9 m

Explanation:

(a)

While the ball is in flight, acceleration due to gravity acts on it. hence

a = acceleration due to gravity = - 9.8 m/s²

The negative sign indicates downward direction

b)

At the maximum height, the ball comes to a momentary stop. hence the velocity of the ball at the maximum height is zero.

0 m/s

c)

Consider the motion of the ball

v₀ = initial velocity of the ball at the time of throw = v

a = acceleration due to gravity = - 9.8 m/s²

t = time interval = 2 s

v = final velocity just before he catch the ball = - v

using the kinematics equation

v = v₀ + a t

- v = v + (- 9.8) (2)

v = 9.8 m/s

d)

h = maximum height reached

v₀ = initial velocity of the ball at the time of throw = 9.8 m/s

v = final velocity of the ball at the highest point = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

using the equation

v² = v₀² + 2 a h

0² = 9.8² + 2 (- 9.8) h

h = 4.9 m

kinetic

i'm so sorry if this is wrong i'm only in 6th grade xd

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