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Displacement in 1-D

A particle is...

Question

A particle is released from a certain height H=400m. Due to wind it gathers a horizontal velocity component v_x=ay where a=√5 s^-1and y is the verticle displacement of particle from the point of release, then find

A) the horizontal drift of particle when it hits the ground.

B) the speed with which particle strikes the ground (g=10m/s²)

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Solution

a) The horizontal drift(velocity) when it strikes the ground would be equal to $v_{x}=\sqrt{5}y=\sqrt{5}\times 400=400\sqrt{5}$ 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 (speed) would be given by: $v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{\left ( 400\sqrt{5} \right )^{2}+\left ( 40\sqrt{5} \right )^{2}}=\sqrt{800000+8000}=\sqrt{808000}$

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