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, [tex]u_y= u\ sin \theta = 60sin(60^0)[/tex]
The magnitude of the displacement after 5s is calculated as;
[tex]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[/tex]
Therefore, the magnitude of the displacement after 5s is 137.31 m.
