Question

A ball of mass 500g is kicked into the air at angle of 45degree. It reaches a height of 12m. What was it's initial velocity?​

Answer 1

Answer:

2 g y = (v sin45)^2

solve for v

v = [sqrt(2gy)]/sin45

Answer 2

$\sqrt{ 240$Answer:

u can work this problem using projectile motion equation or energy conservation.

 projectile motion

hmax=(usin x)squared/2g

12m=(usin45)squared/20

240=(usin45)squared

$\sqrt{240}$=usin 45

u=21.908 m/s

  using energy conservation

KE=PE

1/2m(v)sqr=mgh (When the ball is kicked the kinetic energy is the vertical component of its velocity)

1/2(v)sqr = gh(mass is canceled by mass)

(V)sqr =2gh

         2x 10x12 = 240

V=$\sqrt{240}$ = 15.49 ( but this velocity is one component of the normal velocity which is  vertical component)

     its horizontal component is also the same because sin and cos 45 degree is equal so v=(15.49,15.49)m/s or 21.9 m/s  at angle of 45 to the horizontal.

    Thank you.

Explanation: