Question

An object is thrown vertically and has a speed of 25 m/s when it reaches 1/4 of its maximum height above the ground (assume it starts from ground level). What is the original launch speed of the object?

Answer 1

Answer:

$v_{i} =28.86\frac{ft}{s}$

Explanation:

Conceptual analysis

We apply the kinematic formula for an object that moves vertically upwards:

$(v_{f} )^{2} =(v_{i} )^{2} -2*g*y$

Where:

$v_{f}$ : final speed in ft/s

$v_{i}$ : initial speed in ft/s

g: acceleration due to gravity in ft/s²

y: vertical position at any time in ft

Known data

For $v_{f} = 25\frac{ft}{s}$ ,$y=\frac{1}{4} h$; where h is the maximum height

for y=h,  $v_{f} =0$

Problem development

We replace  $v_{f} = 25\frac{ft}{s}$  , $y=\frac{1}{4} h (ft)$ in the formula (1),

[$25^{2} =(v_{i} )^{2} -2*g*\frac{h}{4}$   Equation (1)

in maximum height(h): $v_{f} =0$, Then we replace in formula (1):

$0=(v_{i} )^{2} - 2*g*h$

$2*g*h=(v_{i} )^{2}$

$h=\frac{(v_{i})^{2} }{2g}$   Equation(2)

We replace (h) of Equation(2) in the  Equation (1) :

$25^{2} =(v_{i} )^{2} -2g\frac{\frac{(v_{i})^{2} }{2g} }{4}$

$25^{2} =(v_{i} )^{2} -\frac{(v_{i})^{2} }{4}$

$25^{2} =\frac{3}{4} (v_{i} )^{2}$

$v_{i} =\sqrt{\frac{25^{2}*4 }{3} }$

$v_{i} =28.86\frac{ft}{s}$