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 = 80.0 above ground level, and the ball is fired with initial horizontal speed . Assume acceleration due to gravity to be = 9.80 .Assume that the cannon is fired at time and that the cannonball hits the ground at time . What is the y position of the cannonball at the time ?
Given that the projectile lands at a distance = 200 from the cliff, as shown in the figure, find the initial speed of the projectile,

Explanation / Answer

acceleration ay =-g velocity vy = -gt + vy0 position y = -gt2/2 + vy0t + y0 acceleration ax = 0 velocity 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, theny(tg) = 0, y(tg) = -gtg2/2 + H = 0 ----> tg = v(2H/g) = 4.04 s The initial speed of the projectile is x(tg) = vx0*tg = vx0*v(2H/g) =200 ==> vx0 = 200/v(2H/g) = 200/(v2*80/9.80) = 49.49 m/s

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