A dummy is fired vertically upward from a cannon with a speed of 40 m/s. How long is the dummy in the air? What is the dummy's maximum height? sdsdsdsdsdsdsdsdsdsd

If you recognize that the graph of height versus time will be a parabola (here air resistance is being ignored, otherwise the graph would be more complex), the symmetry of the curve requires that the slope at time 0 will have the same magnitude and opposite sign as the slope when the dummy returns to the ground. Meaning at impact v = - 40 m/s.

In fact, the velocity is a linear function of time with its slope equal to the acceleration.

v = -9.8 m/s

^{2}* t + 40 m/sAgain the dummy impacts the ground with a velocity of -40 m/s, so you must substitute this value for v and solve for the time of impact to learn how long the dummy was in the air.

-40 m/s = -9.8 m/s

^{2}* t_{Impact}+ 40 m/s-40 m/s - 40 m/s = -9.8 m/s

^{2}* t_{Impact}+ 40 m/s - 40 m/s-80 m/s / (-9.8 m/s

^{2}) = -9.8 m/s^{2}* t_{Impact}/ (-9.8 m/s^{2})t

_{Impact}≈ 8.163 s (If you pay attention to significant digits, this value should be rounded to 8 s since 40 m/s has only one significant digit, but most text books are not careful and will report the answer as 8.2 s or 8.16 s)To find the maximum height, recognize that symmetry of the curve requires that maximum (vertex of the parabola) height occur half way between the time the object left the ground and the time it struck the ground.

t

_{Max Height}= (1/2) * t_{Impact}t

_{Max Height}≈ 4.0816 s (Note that I used the exact time in the calculator rather than the rounded time to calculate this value. You can round the value if the instructor wants this time, but in this problem, the time at the top is not requested. Any rounding will introduce error into your calculation of the maximum height.)Due to the symmetry of the curve, this is also the amount of time required for the dummy to fall from the maximum height to the ground.

The equation for the position of an object undergoing constant acceleration describes the height (y) of the dummy as a function of time.

**y**= (1/2) a *

**t**

^{2}+ v

_{0}*

**t**+ y

_{0}

In this case the equation becomes:

**y**= (1/2) (-9.8 m/s

^{2}) *

**t**

^{2}+ 40 m/s *

**t**+ 0 m

or

**y**= (-4.9 m/s

^{2}) *

**t**

^{2}+ 40 m/s *

**t**

(An alternative approach to solving the problem [or checking the answer] with a graphing calculator is to replace the

**t**with**x**and graph the function. Have the calculator fined the maximum and the x-intercept.)Substitute the value of t

_{Max Height}for t in the equation above to determine the maximum height the dummy reached.y

_{Max}≈ 81.63 mRound this value as required. I prefer to write values with only one significant digit using scientific notation. (e.g. 8. × 10

^{1}m/s)