Question

A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?

Answer 1

Answer:53.63$ms^{-2}$

Explanation:

The equations of motion used in this question is $v=u+at$

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of $9.8ms^{-2}$.

In X-Direction,

Given that initial velocity=$u_{x}$=$33.8ms^{-1}$

Using $v=u+at$,

$v_{x}=33.8+(0)4.25=33.8ms^{-1}$

In Y-Direction,

Given that initial velocity=$u_{x}$=$0ms^{-1}$

Using $v=u+at$,

$v_{y}=0+(9.8)4.25=41.65ms^{-1}$

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$v=\sqrt{1142.44+1734.72}=\sqrt{2877.163}=53.63ms^{-1}$