Question

A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 90.0 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.

Part A

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

Part B

Given that the projectile lands at a distance D = 200 m from the cliff, as shown in the figure, find the initial speed of the projectile, v₀.

Part C

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:

Answer 1

Part A Answer

First solve for t (remember that the cliff is 90m above ground level, so the cannonball has to fall 90m):

y-y₀ = v₀t + 1/2at²
-90 – 0 = 0t + 1/2(-9.8)t²
-90 = -4.9t²
18.3673 = t²

t = 4.2857s

Now we can use the same equation to solve for the height at t/2. Since the problem asks for the height, we’ll use the starting height y₀.

y-y₀ = v₀t + 1/2at²

y-90 = 0(4.2857 / 2) + 1/2(-9.8)(4.2857 / 2)²
y-90 = -4.9(2.1428)²
y-90 = -4.9(4.5918)
y-90 = -22.5

y = 67.5m

Part B Answer

We can solve using the same formula as Part A. The time is 4.2857 seconds (see Part A) and x acceleration is zero.

x-x₀ = v₀t + 1/2at²

200 = v₀(4.2857)

v₀ = 46.7m/s

Part C Answer

y = H – [(gx²)/(2v²_0x)]

Masteringphysics give the above formula so we’ll use it. Since D = 200m, D/2 = 100m:

y = 90 – [(9.8(100)²)/(2(46.7)²)]
y = 90 – (98,000 / 4,355.5804)

One of our readers pointed out that we can also say that this will be the same answer as for Part A, since a projectile travels equal horizontal distances in equal amounts of time.

y = 67.5m