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

v = 25.45 m/s

Step-by-step 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)=(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((h_(max))/(4))=v_o^2-(1)/(2)g((v_o^2)/(g))\\\\v^2=v_o^2-(1)/(2)v_o^2=(1)/(2)v_o^2`

Then, you solve the previous result for vo:

`v_o=√(2)v=√(2)(18m/s)=25.45(m)/(s)`

The initial speed of the object was 25.45 m/s