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.

There is a part 3 however that isn't revealed until part 1 and 2 have been completed. I would appreciate it if you could help me answer that as well after posting the answers and steps on how to obtain the answers to part 1 and 2

Explanation / Answer

1)

First you'll want to find the time it took for the cannonball to drop from the initial height of 50m (Yi=50) to the ground

(Yf=0). Since the ball was shot horizontally, we can assume the vertical velocity is zero (Voy=0) and the acceleration

is 9.8 m/s^2 (gravity)

we plug these numbers into the equation Yf = Yi + Voyt + 1/2at^2

and get 0 = 50 + 0t + 1/2(-9.8)t^2

t will equal5.10 seconds

but you want to know the height at half the time so simply dive t by 2 to get 2.55 s

now

we can solve the height (we'll just call it Y) for time tg/2 by using the same equation. replace the previously

unknown t with the found tg/2 value.
Y = 50 + 0(2.55) + 1/2(-9.8)(2.55^2)
Y =18 m
18 m is your height at tg/2

2)

D=v0*t =160m; H=0.5*g*t^2 =50m;
excluding time t we get:

H=0.5*g*(D/v0)^2, hence v0= D(0.5*g/H) =
= 160(0.5*9.8/50) =50.08m/s;

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