Question

An object is thrown horizontally off of a 37m tall building at a speed of 29 m/s how far away from the building will the object land

Answer 1

Given,

The height of the building, h=37 m

The initial horizontal velocity of the object, u=29 m/s

Given that the object is thrown horizontally. That is the vertical component of the initial velocity of the object is zero, that is u_y=0.

From the equation of motion the height from which the object is given by

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

Where g is the acceleration due to gravity and t is the time it takes for the object to land.

On substituting the known values,

$\begin{gathered} 37=0+\frac{1}{2}\times9.8\times t^2 \\ \Rightarrow t=\sqrt[]{\frac{2\times37}{9.8}} \\ =2.75\text{ s} \end{gathered}$

The range of the object is given by,

$R=ut$

On substituting the known values,

$\begin{gathered} R=29\times2.75 \\ =79.75\text{ m} \end{gathered}$

Thus the object will land 79.75 m away from the building.