Question

A projectile is launched with an initial velocity 60m/s at an angle 60° to the vertical. What the magnitude of it's displacement after 5s.

Answer 1

Answer:

the magnitude of the displacement after 5s is 137.31 m.

Explanation:

Given;

initial velocity of the projectile, u = 60 m/s

angle of projection, θ = 60°

time of motion, t = 5s

the vertical component of the velocity, $u_y= u\ sin \theta = 60sin(60^0)$

The magnitude of the displacement after 5s is calculated as;

$h = u_yt -\frac{1}{2} gt^2\\\\h = 60sin (60^0)\times 5 - \frac{1}{2} (9.8)(5)^2\\\\h = 259.81-122.5\\\\h = 137.31 \ m$

Therefore, the magnitude of the displacement after 5s is 137.31 m.