A cannonball is fired horizontally from the top of a cliff. The cannon is at hei

Question

A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 100 m above ground level, and the ball is fired with initial horizontal speed v0. Assume acceleration due to gravity to be g = 9.80 m/s2 .

Assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg. What is the y position of the cannonball at the time tg/2?

Given that the projectile lands at a distance D = 140 m from the cliff, as shown in the figure, find the initial speed of the projectile, v0.

Explanation / Answer

The acceleration is,

ay =-g
The velocity is,

vy = -gt + vy0

The position is,

y = -gt2/2 + vy0t + y0

Acceleration is,

         ax = 0
velocity is,

     vx = vx0
position x = vx0*t + x0

Now vy0 = 0, y0 = H, giving us

vy = -gt
             y = -gt2/2 + H

If the cannonball hits the ground at tg, then,

y(tg) = 0,
            y(tg) = -gtg2/2 + H = 0
               tg = (2H/g)

y(tg/2) = -g*H/(2*2g) + H
           = -H/4 + H
           =3H/4
           = 300 / 4
           = 75 m

The initial speed of the projectile is,

x(tg) = vx0*tg
             140 = vx0*(2H/g)

The velocity is,

vx0 = 140/(2H/g)
                  = 140/(2*100/9.80)
                  = 30.99 m/s

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