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 = 50.0 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 .

1) 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? Answer numerically in units of meters.

2) Given that the projectile lands at a distance D = 160 m from the cliff, as shown in the figure, find the initial speed of the projectile, v0. Express the initial speed numerically in meters per second.

3)What is the y position of the cannonball when it is at distance D/2 from the hill? If you need to, you can use the trajectory equation for this projectile, which gives y in terms of x directly:

y=Hgx22v20x.

You should already know v0x from the previous part.

Express the position of the cannonball numerically in meters.

Explanation / Answer

(1)
tg = sqrt(2*50/9.8) =3.19 s

tg/2 = 3.19/2 = 1.597 s

y = 50- 1/2 gt^2 = 50-1/2*9.8*1.597^2 = 37.5 m

2) d = v0*tg

V0 = 160/3.19= 50.2 m/s

3)

t = tg/2 , so as in 1 y = 37.5 m

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