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,

$u_x=\frac{R}{t}$

On substituting the known values in the above equation,

$\begin{gathered} 22.2=\frac{36.0}{t} \\ \Rightarrow t=\frac{36.0}{22.2} \\ =1.62\text{ s} \end{gathered}$

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

$y=y_0+u_yt+\frac{1}{2}gt^2$

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,

$\begin{gathered} 0=y_0+0\times1.62+\frac{1}{2}\times-9.8\times1.62^2 \\ y_0=\frac{1}{2}\times9.8\times1.62^2 \\ =12.86\text{ m} \end{gathered}$

Thus the height of the building is 12.86 m