Question

A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 70.0 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. Given that the projectile lands at a distance D = 150 m from the cliff, find the initial speed of the projectile, v0. Express the initial speed numerically in meters per second.

Answer 1

Answer:

39.7 m/s

Explanation:

The motion of the cannonball consists of a horizontal and a vertical motion. The vertical motion is under gravity while the horizontal motion is a constant-velocity motion. Both motions span the same time interval. We consider them differently.

Vertical motion:

Acceleration, $a=9.80$

Initial velocity, $u=0.00$ (since the ball was fired horizontally)

Distance, $s=70.0$

Using the equation $s=ut+\frac{1}{2}at^2$,

$70.0=0+\frac{1}{2}9.80t^2$

$t=\sqrt{\dfrac{70.0}{4.90}}=\dfrac{10.0}{\sqrt{7}}$

Horizontal motion:

Since there is no acceleration, horizontal velocity = horizontal distance ÷ time for vertical motion.

$v=\dfrac{d}{t}$

$v=\dfrac{150}{\frac{10.0}{\sqrt{7}}}$

$v=15\sqrt{7}=39.7$