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A particle is released from a height H= 400m. Due to the wind the particle gathers the horizontal velocity component v x = (sqrt5)y , where y is the vertical displacement of the particle from the point of release, then find: (a) The horizontal drift of the particle when it strikes the ground. (b) the speed with which the particle strikes the ground.

A particle is released from a height H= 400m. Due to the wind the particle gathers the horizontal velocity component vx = (sqrt5)y, where y is the vertical displacement of the particle from the point of release, then find:
(a) The horizontal drift of the particle when it strikes the ground.
(b) the speed with which the particle strikes the ground.

Grade:11

1 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
9 years ago
Dear student,
a) The horizontal velocity when it strikes the ground would be equal tov_{x}=\sqrt{5}y=\sqrt{5}\times 400=400\sqrt{5}
b) The vertical velocity would be given by:
v_{y}^{2}=2gH=2\times 10\times 400=8000 \Rightarrow v_{y}=\sqrt{8000}=40\sqrt{5}
Hence, the net velocity would be given by:
v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{\left ( 400\sqrt{5} \right )^{2}+\left ( 40\sqrt{5} \right )^{2}}=\sqrt{800000+8000}=\sqrt{808000}
Regards
Sumit

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