A 0.250 kg ball is thrown straight upward with an initial velocity of 38 m/s. If air friction is ignored, calculate the:

(a) height of the ball when its speed is 12 m/s
(b) height to which the ball rises before falling
(c) How would your answers to (a) and (b) change if
you repeated the exercise with a ball twice as
massive?

Use energy considerations:

a) 1/2 m vi^2=1/2 mg vf^2+mg*hfinal

b) same equation, vf=0

c) divide the equation in a) by m, and see how the mass then affects that equation.

A ball is thrown straight up from ground level. After a time 4.1 s, it passes a height of 157.1 m. What was its initial speed? The accelera-

tion due to gravity is 9.8 m/s2 . what is its initial speed

To answer these questions, we can use the principles of projectile motion. We'll break down the problem into different parts.

First, let's find the initial velocity when the ball reaches a speed of 12 m/s.

(a) To calculate the height (h1) of the ball when its speed is 12 m/s, we need to find the time it takes for the ball to reach that speed and then use that time to find the height using the formula for vertical displacement.

Step 1: Calculate the time it takes for the ball to reach a speed of 12 m/s.

We will use the equation for final velocity (v) in projectile motion: v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time.

Given: u = 38 m/s, v = 12 m/s, and g = 9.8 m/s².

Rearranging the equation, we have:
t = (v - u) / g

Substituting the given values, we get:
t = (12 - 38) / (-9.8)
t = -26 / -9.8
t ≈ 2.653 seconds (rounded to three decimal places)

Step 2: Calculate the height (h1) of the ball when its speed is 12 m/s.

We will use the formula for vertical displacement in projectile motion: h = ut + (1/2)gt², where h is the vertical displacement.

Given: u = 38 m/s, t = 2.653 seconds, and g = 9.8 m/s².

Substituting the values, we get:
h1 = (38)(2.653) + (1/2)(9.8)(2.653)²
h1 ≈ 81.46 meters (rounded to two decimal places)

Therefore, the height of the ball when its speed is 12 m/s is approximately 81.46 meters.

(b) To find the maximum height (h2) to which the ball rises before falling, we can use the equation for vertical displacement. The maximum height occurs when the velocity of the ball becomes zero.

Given: u = 38 m/s, and g = 9.8 m/s².

We can use the formula for vertical displacement when the final velocity is zero:
0 = u + gt - (1/2)gt²

Rearranging the equation, we get:
(1/2)gt² - ut = 0

Factoring out t, we have:
t((1/2)gt - u) = 0

From this equation, we have two solutions:
t₁ = 0 (representing the initial time when the ball is thrown)
t₂ = (2u) / g (representing the time when the velocity becomes zero)

Substituting the given values, we get:
t₂ = (2)(38) / 9.8
t₂ ≈ 7.755 seconds (rounded to three decimal places)

Now we can calculate the maximum height (h2) using the formula for vertical displacement:

h2 = ut₂ + (1/2)gt₂²

Substituting the values, we get:
h2 = (38)(7.755) + (1/2)(9.8)(7.755)²
h2 ≈ 147.43 meters (rounded to two decimal places)

Therefore, the height to which the ball rises before falling is approximately 147.43 meters.

(c) If the ball were twice as massive, the answers to parts (a) and (b) would not change. The mass of an object does not affect its vertical displacement or its time of flight in projectile motion, assuming that air friction is still ignored.