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,
The magnitude of the displacement after 5s is calculated as;
Therefore, the magnitude of the displacement after 5s is 137.31 m.
Answer:
B. w=12.68rad/s
C. α=3.52rad/s^2
Explanation:
B)
We can solve this problem by taking into account that (as in the uniformly accelerated motion)
( 1 )
where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.
In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have
to calculate the angular speed w we can use
Thus, wf=12.68rad/s
C) We can use our result in B)
I hope this is useful for you
regards