Question

A baseball bat connects with the center of a ball 1.22 meters off the ground, and the ball flies off at a 45 degree angle. It will travel 107 m horizontally before returning to the same height it was when it was initially launched. Does the ball make it over a 7.32 m fence 97.5 m away?

Answer 1

You have to apply the kinematic equations,
\[y=y_0+v_{0y}t-\frac{1}{2}gt^2\]\[x=x_0+v_{0x}t\]\[v_x=v_{0x}\]\[v_y=v_{0y}-gt\]
And\[v_{0y}=v_0\sin\alpha\]\[v_{0x}=v_0\cos\alpha\]
In this case you have to find the initial velocity. To find it employ the range of the ball, 107 m.
\[x=v_0\cos\alpha t\Rightarrow v_0t=x/\cos\alpha \]So
\[0=v_0\sin\alpha t-\frac{1}{2}gt^2\Rightarrow 0=x\tan\alpha-\frac{1}{2}gt^2\Rightarrow t=\sqrt{2x\tan\alpha/g}\]So
\[v_0=\frac{x}{\cos\alpha}\sqrt{\frac{g}{2x\tan\alpha}}\]
where x=107 m, α=45 degrees, g=9.8 m/s^2.

Now to check if the ball will reach the height it ask for, find the time to travel there,
\[t=x_1/(v_0\cos\alpha) \]And calculate the y position at that time,
\[y=y_0+x_1\tan\alpha -\frac{1}{2}g(x_1/(v_0\cos\alpha))^2\]Where, x1=97.5 m, y0=1.22 m. Then compare the result with height given.