Question

Watching a baseball game, you wonder about the ability of the catcher to throw out a base runner trying to steal second base. The catcher is crouched down behind the plate when he observes the runner starting for second. After he gets the ball from the pitcher, he throws as hard as necessary to second base without standing up. If the catcher throws the ball at an angle of 30 from the horizontal so that it is caught at second base at about the same height as that at which the catcher threw it, how much time does it take for the ball to travel the 40 m from the catcher to second base?

Answer 1

\[\overrightarrow{a}=0i-gj\]\[\overrightarrow{v}=|V_o|\cos\theta i+(|V_o|\sin\theta-gt)j\]\[\overrightarrow{s}=|V_o|t\cos\theta i+(|V_o|t\sin\theta-\frac1 2gt^2)j\]by noticing that at the top of the flight the vertical speed component is zero you get\[t'=\frac{|V_o|}{g}\sin\theta\]now notice things like that the total flight time here will be double this, because the final height is the same as the initial.
Now noticing that the horizontal velocity is uniform (because there is no force in that direction) we get the horizontal distance traveled to be\[s_i=|V_o|(2t')\cos\theta=\frac{|V_o|^2}{g}(2\sin\theta\cos\theta)\]which with a little trig looks even nicer:\[s_i=\frac{|V_o|^2}{g}\sin(2\theta)=40\]Since you know the angle you can know the magnitude of the initial velocity. Once you know that you can plug that back into our earlier formula for 2t' to find the total time of flight.