Question

: 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?

Answer 1

Answer:

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

$14.7 - 4.9t^2 = 0$

$4.9t^2 = 14.7$

$t^2 = 14.7/4.9 = 3$

$t = \sqrt{3} = 1.732 s$

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

$r(t) = 8t \hat{i} - 5t\hat{j} + (14.7-4.9t^2)\hat{k}$

$r(1.732) = 8*1.732 \hat{i} - 5*1.732\hat{j} + 0\hat{k}$

$r(1.732) = 13.86 \hat{i} - 8.66\hat{j} + 0\hat{k}$

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

$v(t) = r'(t) = 8\hat{i} - 5\hat{j} - 2*4.9t\hat{k}$

The stone hits the ground at t = 1.732 s

$v(1.732) = 8\hat{i} - 5\hat{j} - 2*4.9*1.732\hat{k}$

$v(1.732) = 8\hat{i} - 5\hat{j} - 17\hat{k}$

The magnitude of this is

$v = \sqrt{v_x^2 + v_y^2 +v_z^2} = \sqrt{8^2 + 5^2 +17^2} = \sqrt{377} = 19.42$m/s