Question

An object is thrown vertically and has an upward velocity of 25 m/s

when it reaches one fourth of its maximum height above its launch point.

What is the initial (launch) speed of the object?

Please explain how to answer this exercise.

Answer 1

The velocity of an object thrown upward decreases with height according to the following law:

$$h=\frac{H_\text{max}}4=\frac{v_f^2-v_i^2}{2g},\\\space\\ v_i^2=v_f^2-2gh=25^2-2·(-9.8)·\frac{H_\text{max}}4.$$

The maximum height:

$$H_\text{max}=\frac{v_f^2-v_i^2}{2g}=\frac{0-v_i^2}{2g}=\frac{-v_i^2}{2g}.$$

Substitute this into the equation above:

$$v_i^2=25^2-2·(-9.8)·\frac{-v_i^2}{4·2·(-9.8)},\\\space\\ v_i^2=\frac{25^2}{1-1/4}=833\text{ (m/s})^2,\\\space\\ v_i=28.9\text{ m/s}.$$