A body of MASS OF 2KG FALLS FREELY FROM REST THROUGH A HEIGHT OF 50M AND COMES TO REST HAVING PENETRATE 5.0CM OF SAND CALCULATE THE TIME TAKING IN FALLING
50 = 1/2 g t^2 ... t = √(100 / 9.8)
Still has to go through the sand though.
It hits the top of the sand in the time R Scott gave you.
How long now until it stops?
assume constant deacceleration in the sand
need speed at sand top
(1/2) m v^2 = m g h
v ^2 = 2 * 9.8 * 50 = 980
v = 31.3 m/s at sand top
average speed during stop = 31.3/2 = 15.7 m/s
time to go 5 cm or 0.05 meters = 0.05 /15.7 = .0032 seconds
not long :)
To calculate the time taken for the body to fall, we can use the equations of motion.
First, let's find the initial velocity (u) of the body as it falls from a height of 50m. We can use the equation:
u^2 = v^2 - 2gh
Where u is the initial velocity, v is the final velocity (which is zero as the body comes to rest), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (50m).
Plugging in the values:
u^2 = 0 - 2 * 9.8 * 50
u^2 = -980
u = -√980 (taking the negative square root because the body is falling)
u ≈ -31.304 m/s
Next, we can use the equation of motion:
v = u + gt
Where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
Plugging in the values:
0 = -31.304 + 9.8 * t
Simplifying the equation:
31.304 = 9.8 * t
t = 31.304 / 9.8
t ≈ 3.19 seconds
Therefore, the time taken for the body to fall is approximately 3.19 seconds.