Question

A stone of mass 5kg fall from top of cliff 50m high and biries 1m in sand. find the average resistance offered by sand and time it take to penitrate.[Ans-2450N, 0.64sec ]​

Answer 1

Answer:

2450,0.64 sec........

Answer 2

Mass of stone = 5 kg

Height of cliff = 50 m

By Conservation of energy, velocity of stone before hitting the ground,

$mgh = \frac{1}{2} m {v}^{2} \\ 5 \times 9.8 \times 50 = \frac{1}{2} \times 5 \times {v}^{2} \\ {v}^{2} = 2 \times 5 \times 98\\ {v}^{2} = 980$

Now the deacceleration. offered by the sand is

${v}^{2} - {u}^{2} = 2as \\ 980 - 0 = 2a(1) \\ 980 = 2a \\ a = \frac{980}{2} \\ a = 490 \: m {s}^{ - 2}$

Since acceleration is opposite to the stone it would be taken as negative,

a = -490 m/s^2

Now, the resistance offered by the sand is

F = m × a

F = 5 × (-490)

F = -2450 N

For the time taken,

$v = u + at \\ 0 = \sqrt{980} + ( - 490)t \\ 490t = \sqrt{980} \\ t = \frac{ \sqrt{490 \times 2} }{490} \\ t = 0.64 \: s$