Question

A balloon starts rising from the ground from rest with on upward acceleration $2m/{{s}^{2}}$. Just after is a stone dropped from it. The time taken by stone to strike the ground is nearly.
A. $0.3s$
B. $1.7s$
C. 1s
D. $1.4s$

Answer 1

Since, a balloon starts rising from ground from rest with an upward acceleration $2m/{{s}^{2}}$ after 1s, and a stone is dropped from it. So we have to use the formula $v=u+at$ and ${{v}^{2}}-{{u}^{2}}=2as$ to find the time taken by stone to strike the ground.

Complete answer:
As we know that when the stone is dropped from the balloon, its initial velocity is the same as the velocity of the balloon at that instant.
The upward acceleration is 2m/s2. The balloon starts from rest, So $u=0$.The balloon rises for 1 sec before the stone is dropped, hence we have

& v=u + at \\\ & v=0+2=2m/s \\\ \end{aligned}$$ In the vertically upward direction. This is the initial velocity of the stone after belonging dropped. The distance moved up by the balloon in 1 second is $$\begin{aligned} & {{v}^{2}}-{{u}^{2}}+2as \\\ & {{v}^{2}}=2as \\\ & s=\dfrac{{{v}^{2}}}{2as}=\dfrac{4}{2\times 2}=1m \\\ \end{aligned}$$ Hence, the stone falls by 1 m before hitting the ground. Now, the acceleration on the store after being dropped is $$g=9.8m/{{s}^{2}}$$ From equation of motion, $$\begin{aligned} & s=ut+\dfrac{1}{2}a{{t}^{2}} \\\ & \Rightarrow 1=-2t+\dfrac{1}{2}9.8\times {{t}^{2}} \\\ & \Rightarrow 4.9{{t}^{2}}-2t-1=0 \\\ & \therefore t=\dfrac{2\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 4.9\times \left( -1 \right)}}{2\times 9.8} \\\ & =\dfrac{2+\sqrt{4+19.6}}{2\times 9.8}=\dfrac{2+\sqrt{23.6}}{9.8} \\\ \end{aligned}$$ Since time can't be the sol $$-ve$$. So, we use the value $$t=\dfrac{2+\sqrt{23.6}}{9.8}$$ $$t=0.7\sec $$ Hence, the stone will strike the ground $$0.75$$ after being dropped from the balloon. **So, the correct answer is “Option B”.** **Note:** Be careful; while using the formula of equation of motion $$v=u+at$$ and $${{v}^{2}}-{{u}^{2}}=2as$$ to find the time taken by stone to strike the ground. Put the value exactly at the same time.