Question

A giant slingshot launches a box of sleeping. Kittens horizontally at 22.2 m/s from the roof of a building and lands 36.0m from the base of the building. How high is the building ?

Answer 1

Given,

The initial horizontal velocity of the projectile, u=22.2 m/s

The total displacement covered by the projectile in the x-direction, i.e., the range of the projectile, R=36.0 m

The projectile was launched horizontally, thus the horizontal component of initial velocity, uₓ=u=22.2 m/s

And the vertical component of the velocity, uy=0 m/s

And the angle of projection is θ=0°

The time interval, t, in which the projectile completes its flight is calculated from the equation,

`u_x=(R)/(t)`

On substituting the known values in the above equation,

`\begin{gathered} 22.2=(36.0)/(t) \\ \Rightarrow t=(36.0)/(22.2) \\ =1.62\text{ s} \end{gathered}`

The height of the building is calculated from the equation of the motion,

`y=y_0+u_yt+(1)/(2)gt^2`

Where y is the final height of the projectile, i.e., y=0 m.

y₀ is the initial height of the projectile, i.e., the height of the building.

g is the acceleration due to gravity.

Let's consider the upward direction as the positive direction. That makes the height of the building a positive value and the acceleration due to gravity a negative value.

On substituting the known values,

`\begin{gathered} 0=y_0+0*1.62+(1)/(2)*-9.8*1.62^2 \\ y_0=(1)/(2)*9.8*1.62^2 \\ =12.86\text{ m} \end{gathered}`

Thus the height of the building is 12.86 m