Question

An object is dropped from rest at a height of 220m and simultaneously another object is dropped from rest at a height 180m. The difference in their heights after 4s if both the objects drop with same acceleration is

Answer 1

Explanation:

Initial difference in height Δh=150m−100m=50m

Distance for each body with time is given by: s=ut+

2

1

at

2

In the above question both the object are released from rest, so initial velocity is: u=0

Since the object are dropped at same time, so decrease in height is,

s

1

=s

2

=

2

1

gt

2

=20 m

Height of first body after 2 s, h

1

=150−20=130 m

Height of second body after 2 s, h

2

=100−20=80 m

Difference in heights =50 m

Answer 2

Answer:

Explanation:

For the first object,

Initial Velocity (u) = 0 m/s

Distance/height = 220 m

Acceleration due to gravity (g)

= 9.8 m/s²

Now, distance travelled in 4s

= ut + 1/2 gt²

= 0×4 + 1/2×9.8×(4)²

= 4.9×16

= 78.4 m

Difference in the height of 1st object = (220-78.4)m = 141.6 m

For the 2nd object,

Initial velocity (u) = 0 m/s

Distance/height = 180 m

Acceleration due to gravity (g)

= 9.8 m/s²

Now, distance travelled in 4s

= ut + 1/2 gt²

= 1/2×9.8×(4)²

= 4.9×16

= 78.4 m

Difference in the height of 2nd object = (180-78.4)m

= 101.6 m

Difference in the height of both = (78.4-78.4) m = 0 m