Question

A projectile is fired at an angle of /4 to the horizontal with an initial speed of 100 m/s.
Find out parametric equations for the projectile motion using time as parameter. Find out the
height of the object after time t=2 sec.

Answer 1

answer

horizontal velocity =$v_x=vcos\theta$

vertical velocity($v_y)=vsin\theta$

vertical displacement

$y=vsin\theta t-\frac{1}{2}gt2$ ........eq.1

horizontal displacement

$x=vcos\theta t$

$t=\frac{x}{vcos\theta}$ .......eq.2

by equation 1 and 2

$y=tan\theta x-\frac{gx^2}{2v^2cos^2\theta}$

this is the parametric equation of projectile.

maximum height can be given as

$h_m=\frac{v^2sin^2\theta}{2g}$

by putting the value

$h_m=\frac{(100)^2sin^24^o}{2(9.81)}=1.83m$