Asked by Esora

A gun is accurately aimed at a dangerous criminal hanging from the gutter of a building. The target is well within the gun’s range, but the instant the gun is fired and the bullet moves with a speed v(sub0) , the criminal lets go and drops to the ground. What happens? The bullet:

1) hits the criminal regardless of the value v(sub0)
2) hits the criminal only if v(sub0) is large enough
3) misses the criminal.

If you were to draw the triangle, width would be 3m, the height would be 2m, and the hypotenuse would be 2^2+3^2=13?

I don't know how to prove that the bullet will hit the criminal with equations. please help! thank you!
Answered by Damon
1)
the bullet accelerates down just as fast as the criminal accelerates down.
Answered by Esora
yes, but how would i prove that with equations?
Answered by Damon
Both are in the air for time t

How far does bullet drop from straight line in time t ?

The only vertical force on the bullet is m g

so it's height as a function of time is
h = (Vo sin T) t - (1/2) g t^2
but Vo sin T t is the straight line to the target ignoring gravity

so it drops (1/2) g t^2

so does the criminal

Answered by bobpursley
both of them move v=gt at the same rate.
Answered by Esora
thank you!
Answered by DIPAYAN
THIS IS OLD... Y'AAL MUST BE OLD BY NOW
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