Question

A cannon, which is 13.5 m away from a 7.9 m high wall, is fired at an angle of 51.9°above the horizontal towards the wall. It takes 1.15 s for the cannon ball to pass a pointthat is directly above the wall.A) What is the speed (in m-s-1) with which the cannon ball was fired?B) What is the vertical distance (in m) by which the cannon ball clears the top of thewall?

Answer 1

Given,

The distance between the cannon and the wall, x=13.5 m

The height of the wall, h₀=7.9 m

The angle of projection, θ=51.9°

The time duration in which the ball passes the point that is directly above the wall, t=1.15 s

A)

The x-component of the initial velocity is given by,

`\begin{gathered} u_x=(x)/(t) \\ \Rightarrow u*\cos \theta=(x)/(t) \\ \Rightarrow u=(x)/(t*\cos \theta) \end{gathered}`

Where u is the initial velocity of the cannon ball.

On substituting the known values,

`\begin{gathered} u=(13.5)/(1.15*\cos51.9^(\circ)) \\ =19.03\text{ m/s} \end{gathered}`

Thus the speed with which the cannon ball was fired is 19.03 m/s

B)

The height of the cannonball at the point that is directly above the wall is given by,

`h=u\sin \theta* t-(1)/(2)gt^2`

Where g is the acceleration due to gravity.

On substituting the known values.

`\begin{gathered} h=19.03*\sin 51.9^(\circ)*1.15-(1)/(2)*9.8*1.15^2 \\ =10.74\text{ m} \end{gathered}`

Thus, the vertical distance by which the cannon ball clears the top of the wall is,

`\begin{gathered} H=h-h_0 \\ =10.74-7.9 \\ =2.84\text{ m} \end{gathered}`

Thus the cannon ball clears the top of the wall by a distance of 2.84 m