At the start of a basketball game, a referee tosses a basketball straight into the air by giving it some initial speed. After being given that speed, the ball reaches a maximum height of 3.60 m above where it started. Using conservation of energy, find: a. The ball's initial speed.
b. The height of the ball when it has a speed of 2.5 m/s.

(a) The ball's initial speed is 8.4 m/s

(b) The height of the ball  is 3.28 m

Explanation:

Given;

maximum height of the ball, H = 3.6 m

Apply conservation of energy;

¹/₂mu² + mgh = ¹/₂mv² + mgH

Where;

m is the mass of the ball

u is the initial velocity of the ball

v is the final velocity of the ball

h is the initial height of the ball

H is the maximum height of the ball

(a) The ball's initial speed, u;

¹/₂mu² + mgh = ¹/₂mv² + mgH

¹/₂u² + gh = ¹/₂v² + gH

make u the subject of the formula

at maximum height, the final velocity  = 0

maximum height H = 3.6 m

initial height, h = 0

g is acceleration due to gravity = 9.8 m/s²

(b) The height of the ball when it has a speed of 2.5 m/s

v = 2.5 m/s

answer:

the answer is d

explanation: