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+(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+(1)/(2)*9.8* t^2 \\ \Rightarrow t=\sqrt[]{(2*37)/(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*2.75 \\ =79.75\text{ m} \end{gathered}`

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