# Stuck on a Problem About Vertical Displacement

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I am not sure how to approach the following question:

A baseball is thrown down a hill. This baseball has on-board sensors that can measure its velocities in both the x- and y- directions. The data show that the ball took off with an initial velocity in the y-direction of 4.0 m/s and had a final velocity in the y-direction of -5.8 m/s . How much lower was the ball when it landed compared to where it started?

I figured out the hang time (1.0 s) based on the V(fy) = v(iy) + at formula. However, how do I get the magnitude of displacement from this? I have the equation for x(max) with the sin and cos -- not going to type it here because I'm not sure how to do thetas and such...in any case, how do I find angle theta based on this information so I can solve the equation?

asked Aug 22, 2017
edited Sep 12, 2018
The vertical velocity at launch is $4m/s$ upwards. When the baseball passes the thrower on its way down its velocity is then $u=4m/s$ downwards. When it lands the baseball's velocity is $v=5.8m/s$ downwards. The vertical distance $s$ of the landing position below the launch position can be found from
$v^2=u^2+2as$.