A stone is thrown vertically upward with a speed of 21.0 m/s

a)How fast is it moving when it reaches a height of 15.8 m

b)How much time is required to reach this height?

Ke = (1/2) m v^2 = (1/2) m (21^2)

gain of potential energy during rise
= m g h = m (9.81)(15.8)

so
Ke at 15.8 = (1/2) m v^2 = (1/2)m(21^2) - m (9.81)(15.8)
m cancels, so compute v at 15.8

v= Vi - g t
so
0.81 t = 21 - v

To find the answers to these questions, we can use the equations of motion for freely falling objects.

The first equation is for calculating the final velocity (v) of an object undergoing constant acceleration:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Since the stone is thrown vertically upward, the acceleration at any point will be the acceleration due to gravity (g), which is approximately -9.8 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity as gravity pulls the stone back down.

a) To find the speed when the stone reaches a height of 15.8 m, we can use the equation:

v^2 = u^2 + 2as

First, let's calculate the initial velocity (u) of the stone. Since the stone is thrown vertically upward, the initial velocity will be positive (+21.0 m/s).

Plugging this into the equation, we have:

v^2 = (21.0 m/s)^2 + 2(-9.8 m/s^2)(15.8 m)

Simplifying this equation will give us the square of the final velocity.

b) To find the time required to reach a height of 15.8 m, we can use the second equation of motion:

s = ut + (1/2)at^2

Rearranging this equation to solve for time (t), we get:

t = (√(2s/a))

Plugging in the values, we have:

t = √(2(15.8 m)/(-9.8 m/s^2))

This will give us the time required to reach a height of 15.8 m.