Suppose a stone is thrown vertically upward from the edge of a cliff on Mars (where the acceleration due to gravity is only about 12 ft/s2 with an initial velocity of 64 ft/s from a height of 192 ft above the ground. The height s of the stone above the ground after t seconds is given by s=−6t2+64t+192
a. Determine the velocity v of the stone after t seconds. b. When does the stone reach its highest point? c. What is the height of the stone at the highest point? d. When does the stone strike the ground? e. With what velocity does the stone strike the ground?

a) v = -12t + 64

b) t = 5.33s

c) s = 362.66ft

d) t = 13.10s

e) v = 93.2ft/s

Explanation:

You have the following equation for the height of a stone thrown in Mars:

      (1)

a)  The velocity of the stone after t seconds is obtained with the derivative of s in time:

  (2)

The equation for the speed of the stone is v = -12t + 64

b) The highest point is obtained when the speed of the stone is zero. Then, from the equation (2) equal to zero, you can obtain the time when the stone is at its maximum height:

The time in which the stone is at the maximum height is 5.33s

c) For this time the stone is at the maximum height. Then, you replace t in the equation (1):

the maximum height is 362.66 ft

d) To find the time when the stone arrive to the ground you equal the equation (1) to zero and you solve for t:

you use the quadratic formula:

You use the result with positive values because is the onlyone with physical meaning.

The time for the stone hits the ground is 13.10 s

e) You replace 13.10s in the equation (2) to obtain the velocity of the stone when it strike the ground:

The minus sign is because the stone's direction is downward.

The speed of the stone just when it strikes the ground is 93.2ft/s

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