A ball which is thrown upward reaches its maximum height
of 100 ft and then returns to the
Starting point. with what speed
was it thrown​

44.27 m/s

Explanation:

Let's set the upwards direction to be positive and the downwards direction to be negative.

Since we do not know the final velocity of the ball right before it hits the ground, we will use our knowledge that the final velocity of the ball is 0 m/s right before it changes direction to fall back to the ground.

Find what constant acceleration equation contains all four variables.

Substitute the known variables into the equation and solve for v₀, the initial velocity.

The initial velocity of the ball 44.27 m/s.

44.27 m/s

Explanation:

Let's set the upwards direction to be positive and the downwards direction to be negative.

Since we do not know the final velocity of the ball right before it hits the ground, we will use our knowledge that the final velocity of the ball is 0 m/s right before it changes direction to fall back to the ground.

a = -9.8 m/s²

Δy = 100 ft

v = 0 m/s

v₀ = ?

Find what constant acceleration equation contains all four variables.

v² = v₀² + 2aΔy

Substitute the known variables into the equation and solve for v₀, the initial velocity.

(0)² = v₀² + 2(-9.8)(100)

0 = v₀² - 1960

1960 = v₀²

v₀ = 14√10 = 44.27188724

The initial velocity of the ball 44.27 m/s.

massxgravity x heght=weightxheight= pe

4900=490xh

h=10