A wooden block is dropped from the top of cliff 100m high and simultaneously a bullet of mass 10g is fired from the foot of the cliff upward with a velocity of 100ms¯¹. the bullet and wooden block will meet each other at what time?

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Solution

$l e t t i s t h e t i m e h i s t h e d i s \tan c e t r a v e l l e d b y b l o c k h = u t + \frac{1}{2} g t^{2} h = \frac{1}{2} g t^{2} d i s \tan c e t r a v e l l e d b y b u l l e t 100 - h = u' t - \frac{1}{2} g t^{2} 100 = u' t t = \frac{100}{100} = 1 s$

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A wooden block of mass 10 g is dropped from the top of a cliff 100 m high. Simultaneously, a bullet of mass 10 g is fired from the foot of the cliff upward with a velocity of $100 m s^{- 1}$ . At what time will the bullet and the block meet ? Take g= $9.8 m s^{- 2}$ .

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A wooden block is dropped from the top of cliff 100m high and simultaneously a bullet of mass 10g is fired from the foot of the cliff upward with a velocity of 100ms¯¹. the bullet and wooden block will meet each other at what time?

let t is the timeh is the distance travelled by blockh =ut+12gt2h = 12gt2distance travelled by bullet100-h =u't-12gt2 100= u'tt = 100100 = 1s

$l e t t i s t h e t i m e h i s t h e d i s \tan c e t r a v e l l e d b y b l o c k h = u t + \frac{1}{2} g t^{2} h = \frac{1}{2} g t^{2} d i s \tan c e t r a v e l l e d b y b u l l e t 100 - h = u' t - \frac{1}{2} g t^{2} 100 = u' t t = \frac{100}{100} = 1 s$