Question

When Babe Ruth hit a homer over the 7.8m -high right-field fence 96m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.2m above the ground and its path initially made a 45 degree angle with the ground. The answer should be in m/s and I do not know how to find the velocity the problem is looking for. Help! Thank you for any suggestions!

Answer 1

This is a projectile motion problem.

Let's realize a few things.
\[y_0 = 1.2 ~ \rm m\]\[y_f = 7.8 ~ \rm m\]\[R = 96 ~ \rm m\]\[\theta = 45^\circ\]

We know, fundamentally, that in a projectile motion problem only one force exists. Gravity. This force only acts in the \(y\) direction.

To begin, let's first break up the initial velocity (which is what we are solving for) into it's \(x\) and \(y\) components. \[v_x = v_o \cos(\theta)\]\[v_y = v_o \sin(\theta)\]

Since no forces act in the \(x\) direction, we know that \(v_x\) is constant. We can easily establish an expression for range if we realize that\[d_x = v_x \cdot t\]where \(d_x\) is the displacement in the x-direction, and \(t\) is the time of flight. We know the ball must travel 96 m, therefore\[R = 96 = v_o \cos(\theta) \cdot t\]

Since gravity causes a constant acceleration, we can use our constant acceleration kinematic equations. \[y(t) = y_0 + v_y \cdot t - {1 \over 2} g t^2\]where \(y(t)\) is the height of the ball at time t, \(y_0\) is the initial height of the ball, \(v_y\) is the initial velocity in the y-direction, g is the acceleration due to gravity, and t is the time of flight.

Here is where things can get tricky. But let's realize that the ball must be 7.8 m above the ground after it has travelled 96 m. Therefore, the initial velocity can be found by solving both the range equation and the equation for \(y(t)\) for \(t\). Let's set that up.

First, solve the range equation for t\[t = {R \over v_0 \cos(\theta)}\]and substitute this into the equation for \(y(t)\)\[y(t) = y_0 + v_0 \sin(\theta) \left [ R \over v_0 \cos(\theta) \right ] - {1 \over 2} g \left [ R \over v_0 \cos(\theta) \right]^2\]We now have one unknown, \(v_0\). We can simplify the above equation a bit to make it more manageable. \[y_f = y_0 + R \tan(\theta) - {g R^2 \over 2v_0^2 \cos^2(\theta)}\]

I'll leave the algebra to you.