Question

A cannon is placed atop a hill that is 120 m high. The cannonball is fired horizontally with an initial velocity of 250 m/s. Find the magnitude of the ball's velocity (in m/s) just before it hits the ground (120 m below the starting point).

Answer 1

Answer:

254.75 m/s

Explanation:

The ball velocity just before it hits the ground would consists of 2 components:

- Horizontal velocity: ignore air resistance, would be the same as initial horizontal velocity, which is 250 m/s

- Vertical velocity: generated by gravitational acceleration g = 10m/s2 after falling for 120m

We can solve for vertical velocity first before it hits the ground using the following equation of motion:

$v^2 - v_0^2 = 2a\Delta s$

where v0 = 0 m/s is the initial vertical velocity of the ball when it's fired, v is the final velocity of the ball when it hits the ground, a = 10 m/s2 is the gravitational acceleration of the ball, and $\Delta s = 120 m$ is the distance traveled.

$v^2 - 0 = 2*10*120 = 2400$

$v = \sqrt{2400} = 49 m/s$

The magnitude of the velocity is the combination of both vertical and horizontal

$v^2 = 49^2 + 250^2 = 2400 + 62500 = 64900$

$v = \sqrt{64900} = 254.75 m/s$