A BODY IS DROPPED FROM A BUILDING 10 M HIGH. CALCULATE THE TIME OF FALL AND THE SPEED WITH WHICH IT HITS THE GROUND.(TAKEg=10M/SECOND SQUARE).
To calculate the time of fall and the speed with which the body hits the ground, we can use the equations of motion.
Let's start by finding the time of fall. We can use the second equation of motion, which relates the distance traveled by an object in free fall to the time taken:
S = ut + (1/2)gt^2
In this equation, S represents the distance traveled, u represents the initial velocity (which is 0 in this case since the body is dropped), g represents the acceleration due to gravity (10 m/s^2), and t represents the time of fall.
Since the body is dropped from rest, the initial velocity (u) is 0. The equation simplifies to:
S = (1/2)gt^2
Substituting the given value for distance (S = 10 m) and the acceleration due to gravity (g = 10 m/s^2) into the equation, we can solve for t:
10 = (1/2)(10)(t^2)
10 = 5t^2
t^2 = 10/5
t^2 = 2
t = √2 ≈ 1.41 seconds
So, the time of fall is approximately 1.41 seconds.
Next, let's calculate the speed with which the body hits the ground. We can use the first equation of motion, which relates velocity to time:
v = u + gt
In this equation, v represents the final velocity, u represents the initial velocity (which is 0), g represents the acceleration due to gravity (10 m/s^2), and t represents the time of fall.
Substituting the known values into the equation:
v = 0 + (10)(1.41)
v = 14.1 m/s
Therefore, the speed with which the body hits the ground is 14.1 m/s.