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:
The location is at (3.535, 1.162) m
Solution:
As per the question:
Mass of the projectile, m = 9.6 kg
Initial velocity, v = 12.4 m/s
Angle,
Mass of one fragment, m = 6.5 kg
Time taken by the fragment, t = 1.42 s
Height of the fragment, y = 5.9 m
Horizontal distance, x = 13.6 m
Now,
To determine the location of the second fragment:
Horizontal Range,
Time of flight, t' =
Now, for the fragments:
Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg
Distance traveled horizontally:
Distance traveled vertically:
Now,
x' = 3.535 m
Similarly,
The location of the other fragment is at (3.535, 1.162)