The height of the rock that is thrown up with a velocity of 13 m/s is given by H = 13t - 1.86*t^2. The velocity of the rock is given by the first derivative of height with respect to time.

`(dH)/(dt)` = 13 - 3.72*t

When the rock hits the...

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The height of the rock that is thrown up with a velocity of 13 m/s is given by H = 13t - 1.86*t^2. The velocity of the rock is given by the first derivative of height with respect to time.

`(dH)/(dt)` = 13 - 3.72*t

When the rock hits the surface its height is equal to 0. Solving 13t - 1.86*t^2 = 0

=> 13t = 1.86t^2

=> t = 6.989 s

The velocity of the rock when it hits the surface is equal to -13 m/s. This the same in magnitude but opposite in direction to the velocity it was thrown at.

**The rock hits the surface after 6.989 s with a velocity -13 m/s**