Question

A balloon carrying a stone rises from rest on the ground with a constant acceleration `10 m//s^(2)`. After 5 s, the stone is released and ultimately it strikes the ground. Sketch a v-t graph for the stone and the maximum height attained by it.

Answer 1

First the stone will accelerate with balloon.
Accelerated motion: `t=0` to `t=5s`
`v=u+a_(1)t=0+10t=50 m//s`
`h_(1)=ut+1/2a_(1)t^(2)=0+1/2xx10xx5^(2)=125 m`
After release, the stone will move under gravity.
Assuming O as the origin and upward direction +ve, O to A through B:
Displacement`=-h_(1)=-125 m`
`-h_(1)=v t_(2)-1/2g t_(2)^(2)`
`-125=50t_(2)-5 t_(2)^(2)`
`t_(2)^(2)-10t-25=0`
`t_(2)=(10+-sqrt(100+100))/2 =(10+-10sqrt(2))/2`
`t_(2)=(10+-10xx1.4)/2 =12 s,-2s`
The stone will strike the ground after 12 s (from its release).
Velocity of the stone at t=12 s,
`v_(0)=v-g t_(2)=50-10xx12=-70 m//s`
v-t graph for stone
A to O: `t=0, v=0`
`t=5 s, v=50 m//s`
`v=10t`
`y=10 x` (straight line)

O to A through B: `v=50-10 t`
`t=0, v=50 m//s`
`t=12 s, v=-70 m//s`
`v=0implies50-10timplies t=5 s`
Shape: y=50-10x(straight line)
y=c+mx
c=50, +ve
m=-10, -ve

Comdining the above two graph, we get the following graph:

The area of the v-t graph from t=0 to t=10 s will give the maximum height attained by the stone.
Maximum height`=1/2xx50xx10=250 m`
OR
If we take time from the start of the balloon, the velocity of the stone in terms of time after its release will be
`v=50-10(t-5)=100-10t (5 s lt = t lt = 17 s)`
`t=5 s, v=50 m//s`
`t=17 s, v=-70 m//s`
`v=0implies100-10T=0impliest=10 s`