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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
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Verified Answer
The correct answers are:
\(T=5 / 2 \mathrm{sec}\)
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 \(v_1=20 \mathrm{ft} / \mathrm{sec}\).
\(S=v_1 t+\frac{1}{2} a_1 t^2\)
\(\Rightarrow-50=20 t+\frac{1}{2}(-32) t^2 \Rightarrow t=\frac{5}{2} \mathrm{sec}\)
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