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In a circus, a joker stands on a highly elevated plank with a ball in his hand. Another joker also stands with a rifle in his hand pointing it directly at the ball. If the fifle is fired precisely at themoment when the ball is released, will the bullet hit the ball ? Air resistance is negligible. |
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Answer» Solution :The gravitational acceleration acting on the bullet and the ball areequal in magnitude and act vertically downwards. As the bullet is firedat the MOMENT the ball is released, the vertical displacements of both the bullet and the ball are the same. Therefore, the bullet will hit the ball. Alternative method : The bullet from the rifle is DIRECTED straight to the ball. Let `v_0` be the velocity with which the bullet leaves the rifle at an angle `theta` with thehorizontal . Let us consider , time taken by the bullet to pass across the vertical line MB at A is t.  Horizontal distance TRAVELLED by the bullet =OB=x [Fig.2.79] `x=v_0cos theta xxt or, t=(x)/(v_0 cos theta)` `therefore t^2=(x^2)/(v_0^2 cos^2 theta)` For thevertical motion of thebullet, `AB=v_0 sin theta xxt-1/2"gt"^2` `=v_0 sin thetaxx(x)/(v_0cos theta)-1/2gxx(x^2)/(v_0^2 cos^2 theta)` `=x tan theta -1/2 (gx^2)/(v_0^2cos^2 theta)` From the`Delta OMB`, `tan theta =(MB)/(OB)=(MB)/x therefore MB=xtan theta` Now, `MA=MB-AB=xtan theta -[xtantheta-1/2*(gx^2)/(v_0^2 cos^2theta)]` `=1/2(gx^2)/(v_0^2cos^2 theta)=1/2"gt"^2` Thus in a time t the bullet falls through a vertical distance `1/2"gt"^2` below M. The vertical distance fallen by the ball is `h=ut+1/2"gt"^2=1/2"gt"^2[because u=0]` Thus the bulletand the ball will always reach the point A at the same time. HENCE the bullet will always hit the ball whatever be the velocity of the bullet. |
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