Trying to find particles speed?

Question

A particle travels between two parallel vertical walls separated by 11m. It moves toward the opposing wall at a constant rate of 6.3m/s. It hits the wall at the same height and the acceleration of gravity is 9.8m/s^2. How would I be able to find the speed of the particle when it hits the opposing wall and then how can I find at what angle with the wall will the particle strike?

Answer 1

The particle is falling as it move across the span between the walls. If it has an initial upward component to it's velocity, that can offset the fall. Now you know the horizontal distance and speed. So it takes;

t = d/vx = 11/6.3 =1.75 seconds to cross the 11 meters

At the opposite wall, the speed the ball goes down with must equal the speed it went up with at teh first wall otherwise it won't hit at the same height. So then:

v = v0 +at ---> -v 0 = v0 -gt where the final speed has a "-" sign bwecause it is point down.

Then: -2v0 = -gt ---> v0 = gt/2 = 9.8*1.75/2 =8.56 m/s

It makes an angle given by : q = arctan(-v0/vx) = arctan(-8.56/6/3) = -53.6 degrees below the hroizontal

Answer 2

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