If a projectile is shot at a velocity of 45 m/s and an angle of pi/6, how far will the projectile travel before landing?

2 Answers
Mar 17, 2018

Range of projectile motion is given by the formula R=(u^2 sin 2 theta)/g where,u is the velocity of projection and theta is the angle of projection.

Given, v=45 ms^-1,theta=(pi)/6

So, R=(45^2 sin ((pi)/3))/9.8=178.95m

This is the displacement of the projectile horizontally.

Vertical displacement is zero,as it returned to the level of projection.

Mar 17, 2018

The projectile will travel =178.94m

Explanation:

The equation of the trajectory of the projectile in the (x,y) plane is

y=xtantheta-(gx^2)/(2u^2cos^2theta)

The initial velocity is u=45ms^-1

The angle is theta=pi/6

The acceleration due to gravity is =9.8ms^-1

When the projectile will land when

y=0

Therefore,

xtantheta-(gx^2)/(2u^2cos^2theta)=xtan(pi/6)-(9.8x^2)/(2*45^2*cos^2(pi/6))=0

x(0.577-0.0032x)=0

x=0.577/0.0032

=178.94m

graph{0.577x-0.0032x^2 [-6.2, 204.7, -42.2, 63.3]}