A cannonball is fired horizontally from the top of a 63 m high cliff. It lands on an invading ship that is 430 m away from the base of the cliff.

At what speed was the cannonball launched?


A) 120 m/s

B) 33 m/s

C) 17.5 m/s

How long does it take to fall 63m?

4.9t^2 = 63
The horizontal speed remains constant, at 430/t m^s

Thank you for helping, but does that mean the answer isn't any of the multiple choices given?

To find the speed at which the cannonball was launched, we can use the principles of projectile motion. Since the cannonball is fired horizontally, its initial vertical velocity is zero and the only force acting on it is gravity. The horizontal distance it travels is 430 m, and the vertical distance (height of the cliff) is 63 m.

First, we can find the time it takes for the cannonball to reach the ship by using the vertical motion equation:

y = v₀y × t + (1/2) × g × t²

Where:
y = 63 m (vertical distance)
v₀y = 0 m/s (initial vertical velocity)
g = 9.8 m/s² (acceleration due to gravity)
t = time

Solving for t, we get:

63 m = (1/2) × 9.8 m/s² × t²

t² = (63 m × 2) / 9.8 m/s²
t² = 12.857

t = √12.857
t ≈ 3.59 s

Next, we can find the horizontal velocity of the cannonball using the horizontal distance and the time it took to reach the ship:

v₀x = d / t

Where:
v₀x = horizontal velocity
d = 430 m (horizontal distance)
t = 3.59 s (time to reach the ship)

v₀x = 430 m / 3.59 s
v₀x ≈ 119.77 m/s

Therefore, the speed at which the cannonball was launched, which is the magnitude of the initial velocity, is approximately 119.77 m/s. Rounded to the nearest decimal, the answer is option A) 120 m/s.