: A stone is thrown from a rooftop at time t = 0 seconds. Its position at time t is given by r(t)=8t i -5t j +(14.7-4.9t^2)k . The origin is at the base of the building, which is standing on flat ground. Distance is measured in meters. The vector i points east, j points north and k points up. A)How high is the rooftop?B)When does the stone hit the ground?C)Where does the stone hit the ground?D)How fast is the stone moving when it hits the ground?

A) 14.7 m

B) 1.732s

C) 13.86m east and 8.66 m south

D) 19.42 m/s

Explanation:

Let x,y,z be the short-cut for east, north and up direction i,j,k, respectively

A) at t = 0, the rock is at the roof top. So the height of the roof top is

z(0) = 14.7 - 4.9*0 = 14.7 m

B) When the stone hits the ground z(t) = 0

C) the position of the stone when it hits the ground is

So it's 13.86m east and 8.66 m south when it hits the ground

D) First we derive the velocity function by differentiate the position function

The stone hits the ground at t = 1.732 s

The magnitude of this is

m/s

be subject to a force andchange its motion