Math Aprd Problem! Velocity!

You throw a ball up in the air from the ground, towards your friend 40 feet above you. (Straight line) The ball is thrown at 56 ft/s, if your friend misses it but catches it on the way down, how long will it have been in the air? 

I believe the equation is something like 0=56t-16t^2

Expert Answers

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Disregarding air resistance etc... the falling body model is `h=-16t^2+v_0t+h_0` where h is the height at time t, `v_0` is the initial velocity, and `h_0` is the initial height. Height is measured in feet and time in seconds.

To be at a height 40 means h=40, `h_0=0` since you throw from the ground, and `v_0=56` as this is the given initial velocity obn the ball.

So `40=-16t^2+56t` or `16t^2-56t+40=0` This can be written as `8(2t^2-7t+5)=0`

Using the quadratic formula we get:

`t=(7+-sqrt(49-4(2)(5)))/(2(2))`

`t=(7+-sqrt(9))/4`

`t=10/4` or `t=4/4`

So the ball is at 40ft at t=1sec and t=2.5sec. Since the ball passed by your friend and he caught it on the way down, the time in the air is 2.5 seconds.

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