**AP Calculus BC Question 62: Answer and Explanation**

### Test Information

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**Question: 62**

**7.** A rock is thrown straight upward with an initial velocity of 50 m/s from a point 100 m above the ground. If the acceleration of the rock at any time *t* is *a* = -10 m/s^{2}, what is the maximum height of the rock (in meters)?

- A. 150
- B. 175
- C. 200
- D. 225

**Correct Answer:** D

**Explanation:**

**D** Because the derivative of velocity with respect to time is acceleration, we have

*v*(*t*) = - 10 *dt* = - 10*t* + *C*

Now we can plug in the initial condition to solve for the constant.

50 = -10(0) + *C*

*C* = 50

Therefore, the velocity function is *v*(*t*) = -10*t* + 50.

Note that the velocity is zero at *t* = 5.

Next, because the derivative of position with respect to time is velocity, we have

*s*(*t*) = (- 10*t* + 50) *dt* = - 5*t ^{2}* + 50

*t*+

*C*

Now, we can plug in the initial condition to solve for the constant.

100 = -5(0)^{2} + 50(0) + *C*

*C* = 100

Therefore, the position function is *s*(*t*) = -5*t*^{2} + 50*t* + 100.

The maximum height occurs when the velocity is zero, so we plug *t* = 5 into the position function to get *s*(5) = -5(5)^{2} + 50(5) + 100 = 225.