Question

An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object

Answer 1

Answer:

v = 25.45 m/s

Explanation:

In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:

$h_{max}=\frac{v_o^2}{g}$   (1)

vo: initial speed of the object = 18 m/s

g: gravitational acceleration = 9.8 m/s²

Furthermore you use the following formula for the final speed of the object:

$v^2=v_o^2-2gh$       (2)

h: height

You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:

$v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2$

Then, you solve the previous result for vo:

$v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}$

The initial speed of the object was 25.45 m/s