question_answer

A balloon starts ascending in vertical upward direction at a constant rate of 5 m/s2. After 2s of its flight begins, a ball has been dropped from it, find the maximum height attained by ball. Also calculate the position of the balloon relative to ground when ball strikes the ground. (Take g = 10 m/s2)

A)  10 m, 25 m        

B)  5 m, 137.1 m

C)  15 m, 56 m        

D)  15 m, 25 m

Correct Answer: C

Solution :

Velocity of balloon at t = 2 s is \[v=0+5\times 2=10m/s\] Height of balloon when ball has been drooped                     \[h=\frac{1}{2}\times 5\times {{2}^{2}}=10m\]from this instant the motion of ball is under gravity, having initial velocity same as that of balloon. Maximum height reached from ground \[=10\times \frac{{{u}^{2}}}{2g}\]\[=10\times \frac{{{(10)}^{2}}}{2\times 10}=15m\] Time taken by ball to reach ground is given by Equation \[s=ut+\frac{1}{2}g{{t}^{2}},\]take up as positive \[\Rightarrow \]\[-10=10\times t-\frac{1}{2}\times 10{{t}^{2}}\]\[\Rightarrow \]\[{{t}^{2}}-2t-2=0\] \[t=2.732,-0.732\]negative time has no meaning. In this time the balloon will reach at a height of \[h=0+\frac{1}{2}\times a\times {{(t+2)}^{2}}\] \[=\frac{1}{2}\times 5\times {{(4.732)}^{2}}\]\[=56m\]