Question

Two tall buildings are 40m apart. With what speed must a ball be thrown horizontally from a window \[145m\] above the ground in one building, so that it will enter a window \[22.5m\] above the ground in another?
$\left( A \right)4m/s$
$\left( B \right)10m/s$
$\left( C \right)8m/s$
$\left( D \right)16m/s$

Answer 1

Hint: The study of motion of the bodies is called kinematics. Apply the kinematic equation of motion, where you substitute the value of $u$ is the initial velocity, $t$ is the time, $g$ is the acceleration of gravity. Apply the kinematic equation in the vertical and horizontal direction. Then solve the above equation to obtain the solution.

Formula used:
$s = ut + \dfrac{1}{2}g{t^2}$
$u$ is the initial velocity, $t$ is the time, $g$ is the acceleration of gravity.

Complete step by step answer:
Motion equation helps to describe a body’s location, velocity or acceleration relative to frame of reference.
The velocity and the position can be derived from the newton equation by the method of integration. Here force acting on a body is known as the function of time.
The study of motion of the bodies is called kinematics. Length means the path between the final and initial position is called length.
Displacement may or may not be equal to the path length travelled of an object.
Distance to unit time is called speed. It is a scalar quantity.
A body is said to be in uniform motion along a straight line when that body is moving with uniform velocity.
Height difference is given by
$s = 145 - 22.5 = 122.5m$
Then apply the kinematic equation
$s = ut + \dfrac{1}{2}a{t^2} - - - - - \left( 1 \right)$
Then along the vertical direction,
$\Rightarrow s > 122.5 = \dfrac{{g{t^2}}}{2} - - - - - - (2)$
Then along the horizontal,
$\Rightarrow 40 = ut$
$\Rightarrow t = \dfrac{{40}}{u} - - - - - \left( 3 \right)$
Solving equation (1), (2) and (3) we get,
Hence, we get
$u = 8m/s$

Hence Option C is the correct option.

Note: The initial velocity will be zero if the motion starts from rest and the frame of reference should be the same. The velocity and the position can be derived from the newton equation by the method of integration. The velocity equation integration results in the distance equation.