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A baseball is hit almost straight up into the air with a speed of 22m/s. Estimate how high it goes and how ling it is in the air
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looks like "almost straight up" should be considered as 90º angle for the initial velocity, i.e. only y-component, in such case the velocity for the ball at any time is: \[v(t)=v_0-gt\] in this case \(v_0=22m/s\) the height of the ball at anytime is given by: \[y(t)=y_0+v_0t-\frac{1}{2}gt^2\] to determine the time it takes to get to the highest point \(H\) we can set the speed to 0, since at that point the ball stops for an instant: \[v_0-gt=0\] with the time we get from that eq. we can calculate \(H\) using the equation for \(y(t)\) give it a try!
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