Question

A balloon is rising vertically upwards with uniform acceleration 2 m/s².
A stone is dropped from it. After 4s, another stone is dropped from it. Find the distance between the two stones 6s after the second
stone is dropped.

first to answer will be marked brainliest

Answer 1

Answer:

The first stone travels for 2+1.5=3.5secs.

The second stone travels for 1.5 secs.

Let first stone be dropped at t=0

Thus, distance travelled by it in 3.5 seconds is 1/2gt ² =4.9(3.5)² =60.025 m

Distance travelled by balloon upwards in 2 seconds= 1/2 at² =0.5×0.2×2 ² =0.4 m

Distance travelled by stone 2 after being released

=0.5×9.8×1.5²

=11.025 m

Thus, distance between the 2 stones at required time

=60.025−(−0.4+11.025)=49.4 m