A ball is projected from the top of tower of height 80 m, with a velocity 30m/s. what is the magnitude of the velocity of the ball when it strikes the ground
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hey dude ur answer is this but this answer is taken from my webisite
The vertical component is under a constant acceleration of g. Whereas the horizontal component remains 10m/s (conservation of momentum) but keeps changing direction soon after it hits the wall. So to reach to the bottom, the ball takes 4 seconds from ( s=ut +0.5 at^2) and it travels 40 m in the horizontal plane (distance) now since the walls are at a distance of 7m every 14 m it travels it reaches on top of the point A. So it reaches A provided the distance is 14x3= 42m since it travels only 40m, the answer should be 2m from A.
Hope you understand.
The vertical component is under a constant acceleration of g. Whereas the horizontal component remains 10m/s (conservation of momentum) but keeps changing direction soon after it hits the wall. So to reach to the bottom, the ball takes 4 seconds from ( s=ut +0.5 at^2) and it travels 40 m in the horizontal plane (distance) now since the walls are at a distance of 7m every 14 m it travels it reaches on top of the point A. So it reaches A provided the distance is 14x3= 42m since it travels only 40m, the answer should be 2m from A.
Hope you understand.
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