Q107 / 155 · WBJEE 2023 · Mathematics
A balloon starting from rest is ascending from ground with uniform acceleration of \(4 \mathrm{ft} / \mathrm{sec}^2\). At the end of \(5 \mathrm{sec}\), a stone is dropped from it. If \(\mathrm{T}\) be the time to reach the stone to the ground and \(\mathrm{H}\) be the height of the balloon when the stone reaches the ground, then
- A\(\mathrm{T}=6 \mathrm{sec}\)
- B\(\mathrm{H}=112.5 \mathrm{ft}\)
- Correct answerC\(T=5 / 2 \mathrm{sec}\)
- D\(225 \mathrm{ft}\)
Solution
Hint : 
Let after \(5 \mathrm{sec}\) ballon is at \(B\) Given: \(a=4 \mathrm{ft} / \sec ^2 u(\) Initial velocity \()=0\) \(\mathrm{t}=5 \mathrm{sec}\) Distance covered by ballon in \(5 \mathrm{sec}\); \(S=u t+\frac{1}{2} \mathrm{at}^2=\frac{1}{2} \times 4 \times 5^2=50 \mathrm{ft}\) Let Speed of balloon at \(B\) be \(\mathrm{V}\) \(V=u+\) at \(=0+4 \times 5=20 \mathrm{ft} / \mathrm{sec}\) So, speed of ball when it's dropped at point \(B\) is \(v_1=20 \mathrm{ft} / \mathrm{sec}\). \(S=v_1 t+\frac{1}{2} a_1 t^2\) \(\Rightarrow-50=20 t+\frac{1}{2}(-32) t^2 \Rightarrow t=\frac{5}{2} \mathrm{sec}\)
Let after \(5 \mathrm{sec}\) ballon is at \(B\) Given: \(a=4 \mathrm{ft} / \sec ^2 u(\) Initial velocity \()=0\) \(\mathrm{t}=5 \mathrm{sec}\) Distance covered by ballon in \(5 \mathrm{sec}\); \(S=u t+\frac{1}{2} \mathrm{at}^2=\frac{1}{2} \times 4 \times 5^2=50 \mathrm{ft}\) Let Speed of balloon at \(B\) be \(\mathrm{V}\) \(V=u+\) at \(=0+4 \times 5=20 \mathrm{ft} / \mathrm{sec}\) So, speed of ball when it's dropped at point \(B\) is \(v_1=20 \mathrm{ft} / \mathrm{sec}\). \(S=v_1 t+\frac{1}{2} a_1 t^2\) \(\Rightarrow-50=20 t+\frac{1}{2}(-32) t^2 \Rightarrow t=\frac{5}{2} \mathrm{sec}\)
Hence, the correct answer is option: 'C'.