Calculus I - Velocity/Distance/Time

[adillhoff]

Homework Statement

In 1939, Joe Sprinz of the San Francisco Seals Baseball Club attempted to catch a ball dropped from a blimp at a height of 800 ft (for the purpose of setting a record).
(a) How long does it take for a ball to drop 800 ft?
(b) What is the velocity of a ball in miles per hour after an 800-ft drop? (88 ft/s = 60 mi/h)
Note: Wind resistance cannot be ignored in this problem. However, even with the slowing effect of wind resistance, the impact of the ball slammed Sprinz's glove hand into his face, fractured his upper jaw in 12 places, broke 5 teeth, and knocked him unconscious. Of course, he dropped the ball.

Homework Equations

h(t) = -1/2g(t)^2 + V(t) + S

The Attempt at a Solution

(a) I am pretty confident I solved this portion correctly. I took the formula h(t) = -1/2g(t)^2 + V(t) + S and plugged in the information I was given:
1. Solved for t: 0 = (-1/2)(32)(t)^2 + (0)(t) + 800
2. 0 = -16(t)^2 + 800
3. 16(t)^2 = 800
4. t^2 = 50
5. t = 7.07 seconds.

(b) Now I have to solve for V, correct? I came up with 154.25 mi/h using the following steps:
1. Take the derivative of the given formula: h(t) = -g(t) + V
2. Plug in the values given: h(7.07) = -32(7.07) + V
3. I set h=0 here to solve for V: 0 = -226.24 + V, or V = 226.24 ft/s
4. Converted to mi/h: 226.24 ft/s * (60 mi/h / 88 ft/s) = 154.25 mi/h.

I am just a little unsure about (b). Can anyone else verify it and let me know if I made any mistakes or not?

 

[tiny-tim]

Welcome to PF!

(b) What is the velocity of a ball in miles per hour after an 800-ft drop? (88 ft/s = 60 mi/h)
…
(b) Now I have to solve for V, correct? I came up with 154.25 mi/h using the following steps:
1. Take the derivative of the given formula: h(t) = -g(t) + V

… = 226.24 ft/s

Hi adillhoff! Welcome to PF!

(a) and (b) are correct, but you should have got (b) directly by using the standard constant acceleration equation, either for u v a and s or u v a and t.

 

[S_Happens]

The original question does mention that wind resistance (drag) cannot be ignored, yet it is not accounted for. Is this a typo, is there missing information in your problem, or are you expected to search this out on your own? Edit- I don't see any reason you'd be considering drag in Calculus I.

In part b, you set the wrong velocity to zero to solve. The derivative of instantaneous position would be instantaneous velocity (which is what you are looking for). The velocity on the right hand side is initial velocity, which you know was 0. Since the ball was always moving in one direction, I myself would have chosen acceleration and velocity to both be in the downward direction to get a positive velocity answer (although either is acceptable when you specifiy direction of velocity). The answer is numerically the same, but you may lose credit.

 

[adillhoff]

Thank you for the swift reply. I believe that the wind resistance part was a typo. There was no information given about wind resistance. I will use your advice and recalculate part B. Thanks again!