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4 July, 06:42

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

+5
Answers (1)
  1. 4 July, 07:03
    0
    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
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