A rocket is traveling upward at a constant velocity of 25m. Att 0, when the rock

Question

A rocket is traveling upward at a constant velocity of 25m. Att 0, when the rocket is 150 m above the ground, a gun at ground level shoots a bullet straight up after the rocket at an initial speed of 75.0m. After being launched, the bullet is in free fall. At t 3.00 s, how far apart are the rocket and the bullet? a) Write an expression for the rocket's height as a function of time, yr(t). Remember to include the initial position, initial velocity, and acceleration. r(t) b) Write an expression for the bullet's height as a function of time, b(t). Remember to include the initial position, initial velocity, and acceleration. m(t) = c) Now write an expression for the distance between the rocket and bullet, (i) d(t), and (ii) solve for the distance att 3.00 s.

Explanation / Answer

(a) An expression for the rocket's height as a function of time, yr (t) which will be given as -

yr (t) = y0 + v0,r t + (1/2) g t2

where, y0 = initial position above the ground

v0,r = initial velocity of rocket

t = time taken

g = acceleration due to gravity

(b) An expression for the bullet's height as a function of time, yb (t) which will be given as -

yb (t) = y0 + v0,b t + (1/2) g t2

where, y0 = initial position above the ground

v0,b = initial velocity of bullet

t = time taken

g = acceleration due to gravity

(c) An expression for the distance between the rocket and bullet which will be given as -

D = [yB (t) - yr (t)]

D = [y0 + v0,b t + (1/2) g t2] - [y0 + v0,r t + (1/2) g t2]

D = v0,b t - v0,r t

D = (v0,b - v0,r) t

At t = 3 sec, we get

D = [(75 m/s) - (25 m/s)] (3 s)

D = [(50 m/s) (3 s)]

D = 150 m

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