A stone is thrown from a rooftop at time t = 0 spends its position at time eleme

Question

A stone is thrown from a rooftop at time t = 0 spends its position at time elementof is given by r(t) = 8ti - 6tj + (20 - 4.9t^2)k The origin is at the Base of the building which is standing on flat grand. Distance is measured in means. The vector i points case, j point North and K points (a) How high is the rooftop above the grand? (b) At what time does the stone hit the grand? (c) How fast is the stone moving what it has the grand? (d) Where does the stone hit the grand? (e) Whats the stones acceleration when it hits the grand? (f) What is the distance the stone traveled? (g) Whats the impact velocity? What it hits the ground? (h) Whats the velocity vector?

Explanation / Answer

given r(t)=8ti -6tj +(20-4.9t2)k

velocity v(t)=r'(t)=8i -6j +(-9.8t)k

acceleration a(t)=r''(t)=0i -0j +(-9.8)k

a)initially(at t=0) stone is at rooftop

height of rooftop above ground = (20-4.9*02)

height of rooftop above ground = 20 meters

b)

when stone hits the ground

20-4.9t2=0

t2=20/4.9

t=2.02 seconds

stone hits the ground after 2.02 seconds

c)

velocity, v(2.02)=8i -6j +(-9.8*2.02)k

velocity, v(2.02)=8i -6j +(-19.8)k

speed =|v(2.02)|

speed =[82+(-6)2+(-19.8)2]

speed =22.18

stone is moving at 22.18 meters per second when it hits the ground

d)

r(2.02)=(8*2.02)i -(6*2.02)j +(20-4.9*2.022)k

r(2.02)=16.16i -12.12j +0k

stone hits the ground at point (16.16,-12.12,0)

e)

when stone hits the ground acceleration ,a(2.02)=0i -0j +(-9.8)k

magnitude of acceleration is 9.8 m/s2

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