Expert: Josh - 3/17/2008

Question
A curveball is pitched with a speed of 64ft/sec.  If the ball is thrown straight upward from a height of 4 feet, find the vertex of the graph of the ball's height and interpret its meaning.

Answer
Hi Justin,

This question is about projectile motion. You can look up this topic on the web, there is a great deal of information out there.

For now, let us work this out from first principle.

The equation we are interested in is x(t)=x(0)+v(0)*t+0.5*a*t^2  ....[1]

Here, x(t) is vertical position of the ball at time t;
x(0) is the initial position;
v(0) is the initial velocity (assuming the ball is thrown vertically upward, this would be the same as its initial speed);
a = 9.8 m^2/s is the gravitational acceleration
(note: we cannot mix SI units with imperial units. So, let's convert the speed from ft/sec to m/sec, or vice versa, if you are from Great Britain.)

Conversion: 1 [inch] = 2.54 [centimeter]
1 [feet] = 12 inches = 12 x 0.0254 = 0.3046 [meter]
Thus, v(0) is approximately 19.5072 [m/sec].

Taking the first derivative of the distance equation in [1], we obtain the velocity equation: dx/dt = v(t) = v(0)+a*t ....[2]
(I'm not sure if you have learned this in maths or physics...but it's a good thing if you know about differentiation. Otherwise you need to commit this to memory)

At the highest position, the ball must reach an instantaneous speed of zero, since it changes direction, i.e., transitions from an upward to downward trajectory. Since we are looking at the motion path from the side, in one dimension velocity (along the vertical axis) and speed are basically the same thing.

To find the time when the ball reaches the highest position, we set the velocity dx(t)/dt = v(t) = 0 and solve for t. The ball must momentarily come to rest before it turns around and fall back down to earth.

i.e., solving for 0 = v(0) + a*t, we get t=v(0)/a, where initial velocity v(0)=19.5072 [m/sec] and gravitational acceleration "a"=9.8 [m/sec^2]. Thus, the time taken to rise to the highest point is t = 2 sec.

Recall the initial height is x(0) = 4 x 0.3046 = 1.2192 [meter].
Substituting t=2 into equation [1] gives you the maximum height.
i.e., you evaluate x(t)=x(0)+v(0)*t+0.5*a*t^2 at t=2.

Now, knowing the time when the peak is reached "t_max", and the maximum height "x_max", you have the vertex (t_max,x_max), where x_max measures the distance along the vertical axis.