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
[tex]14.7 - 4.9t^2 = 0[/tex]
[tex]4.9t^2 = 14.7[/tex]
[tex]t^2 = 14.7/4.9 = 3[/tex]
[tex]t = \sqrt{3} = 1.732 s[/tex]
C) the position of the stone when it hits the ground is
[tex]r(t) = 8t \hat{i} - 5t\hat{j} + (14.7-4.9t^2)\hat{k}[/tex]
[tex]r(1.732) = 8*1.732 \hat{i} - 5*1.732\hat{j} + 0\hat{k}[/tex]
[tex]r(1.732) = 13.86 \hat{i} - 8.66\hat{j} + 0\hat{k}[/tex]
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
[tex]v(t) = r'(t) = 8\hat{i} - 5\hat{j} - 2*4.9t\hat{k}[/tex]
The stone hits the ground at t = 1.732 s
[tex]v(1.732) = 8\hat{i} - 5\hat{j} - 2*4.9*1.732\hat{k}[/tex]
[tex]v(1.732) = 8\hat{i} - 5\hat{j} - 17\hat{k}[/tex]
The magnitude of this is
[tex]v = \sqrt{v_x^2 + v_y^2 +v_z^2} = \sqrt{8^2 + 5^2 +17^2} = \sqrt{377} = 19.42[/tex]m/s
