Question

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how far would the cannonball travel before it lands on the ground? (Show all work)

Answer 1

Answer:

The cannonball travels a distance of 26.46 m before it lands on the ground.

Explanation:

Initial horizontal velocity of the cannon ball = 48.1 m/s

The initial vertical component of the velocity is zero

Time taken in reaching the ground by travelling 1.5 m distance in the vertical direction

Using second equation of motion

$h=ut+\frac{1}{2}gt^2$

$1.5=0\times t+\frac{1}{2}\times 10\times t^2$

$\implies t^2=\frac{3}{10}$

$\implies t=\sqrt{\frac{3}{10}}$

$\implies t=0.55$ second

There is no acceleration in the horizontal direction

Thus, in 0.55 sec, the distance covered by the cannon

=horizontal velocity × time

$=48.1\times 0.55$

$=26.455$ m

Hope this helps.