A wooden block of mass `10 g` is dropped from the top of a tower `100 m` high. Simultaneously, a bullet of mass `10 g` is fired from the foot of the tower vertically upwards with a velocity of `100 m//s`. If the bullet is embedded in it, how high will the block rise above the top of tower before it starts falling? `(g=10m//s^2)` A. `75m`B. `85m`C. `80m`D. `10m`

A wooden block of mass `10 g` is dropped from the top of a tower `100

1.	A wooden block of mass `10 g` is dropped from the top of a tower `100 m` high. Simultaneously, a bullet of mass `10 g` is fired from the foot of the tower vertically upwards with a velocity of `100 m//s`. If the bullet is embedded in it, how high will the block rise above the top of tower before it starts falling? `(g=10m//s^2)` A. `75m`B. `85m`C. `80m`D. `10m`
Answer» Correct Answer - A The bullet and block will meet after tme `t=h/(u_(rel))=100/100=1` During this time, distance travelled by the block, `s_(1)=1/2"gt"^(2)=1/2xx10xx1^(2)=5m` Distance travelled by the bullet, `s_(2)=100-s_(1)=96m` Velocity of the bullet before collision `u_(2)=u-"gt"=100-10xx1=90m//s` Velocity of the block before collision. `u_(1)"gt"=10m//s` Let `V` be the combined velocity after collision. According to the law of conservation of momentum. `m_(1)u_(1)+m_(2)u_(2)=(m_(1)+m_(2))V` (Velocity in upward direction in considered positive) Solving we get `V=40m//s` Maximum height risen by the block `=(V^(2))/(2g)=80m` Height reached above the top of the tower is `80-s_(1)=80-5=75m`

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