Question

A stone is thrown vertically upward with a speed of 29.0 m/s and when it reaches a height of 13 m, the velocity is 24.2 m/s.

Answer 1

Answer:

the value of t = 0.49 seconds shows that its upward journey

and

at t = 5.43 seconds shows in downward journey

Explanation:

Given:

initial speed, u = 29 m/s

acceleration due to gravity, g = - 9.8 m/s^2

h = 13 m

Let it is moving with velocity v at a height of 13 m.

Use third equation of motion

v² = u² + 2gh

By substituting the values

v² = 29² - (2 * 9.8 * 13)

v = sqrt 585.94

v  = 24.2 m/s

Let it takes time t to reach at height 13 m

Use second equation of motion

s = u * t + 1/2 * g * t²

13 = 29t - 4.9t²

4.9t² - 29t + 13 = 0

using quadratic equation to solve time

     29 ± $\sqrt{29^2 - 4 * 4.9 * 13}$

t = ------------------------------------

                     2 * 4.9

t = 5.43 second or t = 0.49 second

Therefore...

the value of t = 0.49 seconds shows that its upward journey

and

at t = 5.43 seconds shows in downward journey