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 4.1 m ,. neglect air resistance. answer in units of m/s

Answer 1

The initial vertical speed of a ball thrown upward to reach a maximum height of 4.1 meters is approximately 8.97 m/s. This is calculated using the kinematic equation for vertical motion under gravity, considering that final velocity at the maximum height is zero and the acceleration due to gravity is -9.8 m/s².

Step-by-step explanation:

To find the initial vertical speed of a ball thrown upwards given the maximum height, we can use the kinematic equation for vertical motion under gravity:

vf² = vi² + 2a * d

Where vf is the final velocity, vi is the initial velocity, a is the acceleration due to gravity, and d is the maximum height. At the maximum height, the final velocity vf is 0 m/s, the acceleration due to gravity a is -9.8 m/s2, and d is 4.1 m.

Plugging in the values, we get:

02 = vi² + 2(-9.8) * 4.1

vi² = 2(9.8) * 4.1

vi² = 80.36

vi = √80.36

vi = 8.97 m/s (approximately)

The initial vertical speed of the ball is approximately 8.97 m/s upwards.

Answer 2

the question is missing another parameter. we need to know the time taken for the ball to come down. You can use the below formula to find the initial vertical speed.

S = 1/2at2 + ut

distance = 1/2 × acceleration × time2 + intial speed × time