**Applied Calculus Questions and Solutions**

We throw a ball upwards. Its height *h* (in meters), *t* seconds after its launch is given by the following equation

*h*(*t*) = –*t*^{2} +50*t*+1

- Find the initial height of the ball.
**Round your answer to 4 decimals if needed and show your calculations. Use the correct notation.** - Find at what time the speed of the ball is zero and find the height of the ball at this moment.
**Round your answer to 4 decimals if needed and show your calculations. Use the correct notation.** - Use the derivative to find the speed of the ball precisely when it will touch the ground after its launch.
**Round your answer to 4 decimals if needed and show your calculations. Use the correct notation.** - Calculate the speed and the acceleration of the ball 3 seconds after its launch.
**Round your answer to 4 decimals if needed and show your calculations. Use the correct notation.** - Interpret the results of c in the context of the problem. You must write an interpretation ….

**Solutions**

**a) Initial height of the ball**

Initial height is when time t = 0

*h*(*t*) = –*t*^{2} +50*t*+1

*h* = -0^{2} +50 x 0+1 = 1

*h* = 1 meter

**b) Time at which speed of the ball is zero, and height of the ball at this moment**

Speed s = …..

Solutions: t = 25 seconds, *h* = 626 meters

Find all worked out (well explained) solutions to the above questions.

(e): Interpretation for question c:

In this case, time cannot be negative and so the negative root of the quadratic equation does not apply. Negative speed indicates the ball is moving downwards (and not upwards as was the case when the ball started moving). The acceleration of the ball is constant from start to finish.