Section 3.3 Problem 8

If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is given by:

position[t_]  =  80 t  -  16 t^2

A plot of the position function looks like this:

Plot[position[t], {t, 0, 5},  
      AxesLabel ->  {"Time (s)", "Position (ft)"}]

We see that the ball initially rises and then returns to earth in a parabolic shape. From the graph it appears that the maximum height the ball reaches is 100 ft, but we need to prove this, not read it from a graph!

To find the maximum height the ball reaches, we need to know the time that the ball reaches the maximum height. When the ball is at its maximum height, the velocity of the ball will be zero. The velocity of the ball is given by the derivative of the position function:

velocity[t_]  =  80  -  32 t

Setting the velocity equal to zero and solving for the time gives us the time when the ball is at its maximum height:

timemax  =  Solve[velocity[t]  ==  0, t]

The time the ball is at its maximum is at 5/2, and the position at that time is:

position[5/2]

The ball therefore reaches a maximum height of 100 ft, as we had originally guessed from the graph.

To find the velocity of the ball when the ball is at 96 ft above the ground, we need to find out at what times the ball is at 96 ft. From the graph, we see that this occurs at two times, when the ball is going up and when the ball is coming down:

Plot[{position[t], 96}, {t, 0, 5},  
      AxesLabel ->  {"Time (s)", "Position (ft)"}]

To solve for these two times, we must solve the equation:

eq1  =  position[t]  ==   96

Solve[eq1, t]

Thus, at time 2 s the ball is rising and at 96 ft, and at time 3 s the ball is falling and is at 96 ft. The velocity for these two cases is found from evaluating the velocity at these times:

velocity[2]

velocity[3]

Therefore, the velocity is 16 ft/s when the ball passes 96 ft on the way up, and -16 ft/s when the ball passes 96 ft on the way down.