a) 2.75 s
The vertical position of the ball at time t is given by the equation

where
h = 4 m is the initial height of the ball
u = 12 m/s is the initial velocity of the ball (upward)
g = 9.8 m/s^2 is the acceleration of gravity (downward)
We can find the time t at which the ball reaches the ground by substituting y=0 into the equation:

This is a second-order equation. By solving it for t, we find:
t = -0.30 s
t = 2.75 s
The first solution is negative, so we discard it; the second solution, t = 2.75 s, is the one we are looking for.
b) -15.0 m/s (downward)
The final velocity of the ball can be calculated by using the equation:

where
u = 12 m/s is the initial (upward) velocity
g = 9.8 m/s^2 is the acceleration of gravity (downward)
t is the time
By subsisuting t = 2.75 s, we find the velocity of the ball as it reaches the ground:

And the negative sign means the direction is downward.
Answer:
v = 9.936 m/s
Explanation:
given,
height of cliff = 40 m
speed of sound = 343 m/s
assuming that time to reach the sound to the player = 3 s
now,
time taken to fall of ball


t = 2.857 s
distance
d = v x t
d = v x 2.875
time traveled by the sound before reaching the player



distance traveled by the wave in this time'
r = 0.143 x 343
r= 49.05 m
now,
we know.
d² + h² = r²
d² + 40² = 49.05²
d =28.387 m
v x 2.875=28.387 m
v = 9.936 m/s
The correct answer is hook shot
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