A cannon fires a 0.652 kg shell with initial

velocity vi = 12 m/s in the direction = 61 ◦

above the horizontal.

The shell's trajectory curves downward be-

cause of gravity, so at the time t = 0.473 s

the shell is below the straight line by some

vertical distance h.

Find this distance h in the absence of

air resistance. The acceleration of gravity is

9.8 m/s2.

Answer in units of m

Mass have no effect for the projectile motion and u want to know the height "h"

first,

find the vertical and horizontal components of velocity

vertical component of velocity = 12 sin 61

horizontal component of velocity = 12 cos 61

now for the vertical motion;

S = ut + (1/2) at^2

where

s = h

u = initial vertical component of velocity

t = 0.473 s

a = gravitational deceleration (-g) = - 9.8 m/s^2

h=[12*sin 610*0.473]+[-9.8 * (0.473) 2]

u can simplify this and u will get the answer

h=.5Gt2

H=1.09m