Question

A ball is thrown upward. What is its initial vertical speed? The acceleration of gravity is 9.8 m/s^2 and maximum height is 4.3 m. Neglect air resistance.. Answer in units of m/s.

Answer 1

To find the initial vertical speed of a ball thrown upward that reaches a maximum height of 4.3 m, we use the kinematic equation and solve for the initial velocity. The initial vertical speed is found to be approximately 9.17 m/s.

Step-by-step explanation:

The question involves calculating the initial vertical speed of a ball thrown upward given the maximum height it reaches. To solve this, we use the kinematic equation that relates the initial velocity, acceleration due to gravity, and the maximum height a projectile reaches.

Here's the kinematic equation that we'll use:

vf^2 = vi^2 + 2a * d

Where:

• vf is the final velocity (0 m/s at the maximum height)
• vi is the initial velocity
• a is the acceleration due to gravity (-9.8 m/s^2, it's negative because gravity is pulling the ball down)
• d is the maximum height reached (4.3 m)

Solve for vi (initial velocity):

0 = vi^2 - 2(9.8 m/s^2)(4.3 m)

vi^2 = 2(9.8 m/s^2)(4.3 m)

vi = √(2 * 9.8 m/s^2 * 4.3 m)

vi = √(84.14 m^2/s^2)

vi ≈ 9.17 m/s

The initial vertical speed of the ball is approximately 9.17 m/s.

Answer 2

h = v o · t - g t² / 2
v = v 0 - g t
0 = v o - 9.8 t
v o = 9.8 t
4.3 = 9.8 t · t - 9.8 t² / 2
4.3 = 9.8 t² - 4.9 t²
4.3 = 4.9 t²
t² = 4.3 : 4.9
t² = 0.87755
t = √ 0.87755
t = 0.93678 s
v o = 9.8 · 0.93678
v o = 9.18 m/s
Answer:
its initial vertical speed is 9.18 m/s.