Question

3. A balloon starts rising from the ground with an acceleration of 1.25 m/s2. A stone is released from the balloon after 10s. Determine (1) maximum height of stone from ground (2) time taken by stone to reach the ground

NEET/Medical Exams

Physics

Solution

323

4.0 (1 ratings)

[ begin{array}{l} c^{n} d 1 cdot 25 m / s^{2} t=8 s end{array} ] ( S(text { distance } operatorname{traveled} operatorname{in} 88)=frac{1}{2}(8)^{2}(1-25) ) [ s=40 mathrm{m} ] ( begin{aligned} v(text { velocify after } 8 s) &=(1.25)(8) &=4 n / s end{aligned} ) For stone [ S=u t+frac{1}{2} a t^{2} ] ( Rightarrow-40=10(t)-frac{1}{x}left(t^{5}right) t^{2} ) a ( frac{5 t^{2}-10 t-40=0}{frac{t^{2}-2 t-8}{5}=0}+begin{array}{l}t=-2,4 x_{1}end{array} ) ( therefore sqrt{t}=4 ) Q maxm height of etome fuom ground = ( 90 mathrm{m} ) (11) tine laker loy stowe to resen groud = 4 sec