Answer:
A.) 4.52s
B.) 44.3 m/s
Explanation:
Given that a cargo of mass 50kg fall freely from rest at a height of 100m and came to rest having penetrated 5.0cm of the ground
The total height = 100 + 5/100
The total height = 100 + 0.05
The total height = 100.05 m
Since the cargo fall from rest, the initial velocity U will be equal to zero. That is,
U = 0
Use second equation of motion
h = Ut + 1/2gt^2
Substitutes g = 9.8 and others parameters into the formula
100.05 = 0 + 1/2 × 9.8 × t^2
100.05 = 4.9t^2
t^2 = 100.05/4.9
t^2 = 20.42
t = sqrt( 20.42)
t = 4.52 s
B.) To calculate the velocity of the body when it hits the ground,
You can achieve it by first calculating its kinetic energy. Or by using the first equation of motion.
So, let use first equation of motion
V = U + at
But U = 0
Substitute the time and g into the formula
V = 9.8 × 4.52
V = 44.29 m/s
Therefore, the velocity of the body when it hits the ground is 44.3 m/s approximately.
