Question

How can the initial velocity of a ball be determined

Answer 1

solution

let us assume that a ball is projected with the initial velocity u making an angle $\theta$ with the horizontal X axis.

then ball will perform projectile motion in xy plane as shown in figure.

now

initial velocity have two component first in +ve x-axis and second in +ve y axis.these component can be written in terms of $\theta$ by

using trignometry

$u_x=ucos\theta$

and $u_y=usin\theta$

during the motion of ball u_x remain constant because there are no any accelaration in the direction of x axis motion but u_y will very because accelaration due to gravity is working in -ve y axis direction.

so when ball will be at the maximam height during its motion , the velocity of ball in verticle direction will be zero or we can say it will have only horizontal velocity.

by applying the newton's equation of motion in verticle direction

we will get

$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$ .........eq.1

at maximum height $\overrightarrow{v}=0m/s$

$\overrightarrow{a}=-g\space m/s^2$ (g is accelarationn due to gravity)

$\overrightarrow{u}=usin\theta$

then putting these value in equation 1

$0=usin\theta -gt$

$usin\theta=gt$

$\fcolorbox{red}{aqua}{$u=\frac{gt}{sin\theta}$}$

if t and $\theta$ are known then initial veloccity can be calculated.

by applying newtons third equation of motion in verticle direction

v_y²=u_y²+2as ..............eq.2

in verticle direction at maximum height

v_y=0 m/s

$a=-g\space m/s^2$

$u_y=usin\theta$

$s=h_{max}$

by putting these value in equation 2

we get

$0=(usin\theta)^2 -2gh_{max}$

$u^2=\frac{2gh_{max}}{sin^2\theta}$

$\fcolorbox{red}{aqua}{$u=\sqrt\frac{2gh_{max}}{sin^2\theta}$}$

if maximun height and $\theta$ is known then initial velocity can be calculated by above equation.