A ball is thrown vertically upward with a speed of +10.0 m/s.

(a) How high does it rise?
m

(b) How long does it take to reach its highest point?
s

(c) How long does the ball take to hit the ground after it reaches its highest point?
s

(d) What is its velocity when it returns to the level from which it started?
m/s

Someone will be glad to critique your thinking. For using conservation of energy for the first question.

For (b), require that the ball decelerate at rate g until the vertical velocity is zero.

To answer these questions, we can use the kinematic equations of motion. Let's break down each part to get the answer.

(a) To find how high the ball rises, we need to use the equation:

vf^2 = vi^2 + 2ad

Where:
vf is the final velocity (0 m/s at the highest point),
vi is the initial velocity (+10 m/s),
a is the acceleration due to gravity (-9.8 m/s^2),
and d is the displacement (the height we want to find).

Rearranging the equation, we have:

d = (vf^2 - vi^2) / (2a)

Plugging in the values, we get:

d = (0^2 - 10^2) / (2*(-9.8))
= (-100) / (-19.6)
= 5.1 m (rounded to one decimal place)

Therefore, the ball rises to a height of 5.1 meters.

(b) To find the time it takes to reach the highest point, we can use the equation:

vf = vi + at

Since the ball is at its highest point, the final velocity (vf) is 0 m/s. The initial velocity (vi) is +10 m/s, and the acceleration (a) is -9.8 m/s^2 (due to gravity).

Rearranging the equation, we have:

t = (vf - vi) / a

Plugging in the values, we get:

t = (0 - 10) / (-9.8)
= 1.02 s (rounded to two decimal places)

Therefore, it takes approximately 1.02 seconds to reach the highest point.

(c) To find the time it takes for the ball to hit the ground after reaching its highest point, we can use the equation:

d = vit + (1/2)at^2

Where:
d is the displacement,
vi is the initial velocity (0 m/s at the highest point after reaching its peak),
a is the acceleration due to gravity (-9.8 m/s^2),
and t is the time we want to find.

Since the displacement is equal to the height it rises (5.1 m), and the initial velocity is 0 m/s, we have:

5.1 = 0t + (1/2)(-9.8)t^2

Rearranging the equation, we get:

4.9t^2 = 5.1

t^2 = 5.1 / 4.9
t^2 = 1.04

Taking the square root of both sides, we get:

t = √1.04
t = 1.02 s (rounded to two decimal places)

Therefore, it takes approximately 1.02 seconds for the ball to hit the ground after reaching its highest point.

(d) When the ball returns to the level it started, its velocity will be equal to the initial velocity. In this case, the initial velocity is +10 m/s.

Therefore, the velocity of the ball when it returns to the level from which it started is +10 m/s.