A body is dropped from rest at a height of 150 m and simultaneously another
body is dropped from rest from a point 100 m above the ground. What is the
difference between their heights after they have fallen for 3 seconds?
1st Body:
h1 = ho - 0.5g*t^2 = 150 - 4.9*3^2 = 105.9 m. above gnd.
2nd Body:
h2 = ho - 0.5g*t^2 = 100 - 4.9*3^2 = 55.9 m. above gnd.
h1 - h2 = 105.9 - 55.9 = 50 m. = Difference between their hts.
To find the difference in height between the two bodies after 3 seconds, we need to determine the distance each body has traveled in that time.
Let's start by calculating the distance traveled by the first body, which is dropped from a height of 150 m. We can use the equation of motion for an object in free fall:
d = (1/2) * g * t^2
Where:
d is the distance traveled
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time (in seconds)
For the first body:
d1 = (1/2) * 9.8 * (3^2)
= 44.1 m
Next, we calculate the distance traveled by the second body, which is dropped from a height of 100 m:
d2 = (1/2) * g * t^2
= (1/2) * 9.8 * (3^2)
= 44.1 m
The difference in height between the two bodies after 3 seconds is the initial difference in height minus the distance traveled by each body:
Difference = (150 - 100) - (d1 - d2)
= 50 - (44.1 - 44.1)
= 50 - 0
= 50 m
Therefore, the difference in height between the two bodies after 3 seconds is 50 meters.