A rocket is dropped out of a window and pointed sideways toward another building

Question

A rocket is dropped out of a window and pointed sideways toward another building. One second after it is dropped, its motor fires, giving it an accleration of 2g in the horizontal direction. (Its vertical accleration is still g downward, so you'd have to sum them.) After the motor fires, the rocket flies along its new path until it strikes another building, located 40m away.

1. How long (in total) was the rocket in the air?

2. How far below the level of the window did the rocket strike the building?

Explanation / Answer

a)

here by using the second equation of motion

x = u*t + 0.5 * a * t^2

a = 2g = 2 * 9.8 = 19.6 m/s^2

u = 0 m/s

40 = 0 + 0.5 * 19.6 * t^2

t = sqrt( 4.08)

t = 2 sec

one second after it is droped so the total time is = 3 sec

b)

here we use the same formula

y = 0 + 0.5 * a * t^2

y = 0.5 * 9.8 * 3 * 3

y = 44.1 m

so the rocket strike the building 44.1 below the window

if you use a = 10m/s^2 the answer will change then its answer is 45 m

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