Question

A child throws a water filled balloon at an angle 60 with a speed 10m/s . A car is advancing towards the child with speed 10m/s . If the balloon has to hit the car, how far away should the car be when the balloon is to be thrown??

Answer 1

Solution:-The balloon will meet the car at the distance at the point B

Time taken by balloon to reach AB=

$\frac{2usin \alpha }{g} = \frac{2 \times 10 \times sin60}{g} = \frac{2 \times 10 \times 1}{20} \\ = 1s$

Distance travelled by man AB=

$\frac{2 {u}^{2} sin \alpha cos \alpha }{g} = \frac{2 \times 100sin60cos60}{10} \\ = 2 \times 10 \times \frac{ \sqrt{3} }{2} \times \frac{1}{2} = 5 \sqrt{3}m \\ time \: taken \: by \: man \: to \: reach \: at \: B \: = tim \: taken \\ by \: balloon \: to \: reach \: from \: A \: to \: C$

$\frac{2 {u}^{2} sin \alpha cos \alpha }{g} = \frac{2 \times 100sin60cos60}{10} \\ = 2 \times 10 \times \frac{ \sqrt{3} }{2} \times \frac{1}{2} = 5 \sqrt{3}m \\ time \: taken \: by \: man \: to \: reach \: at \: B \: = tim \: taken \\ by \: balloon \: to \: reach \: from \: A \: to \: C = 1s \\ so \: BC = 10 \times 1 = 10m \\ distance \: of \: balloon \: from \: car = 5 \sqrt{3} + 10 \\ = 8.65 + 10 = 18.65m$

hence balloon should be at the distance of 18.65m from the car before it get thrown

{hope it helps}