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anastassius [24]
3 years ago
5

A 500 g model rocket is resting horizontally at the top edge of a 40-m-high wall when it is accidentally bumped. The bump pushes

it off the edge with a horizontal speed of 0.5 m/s and at the same time causes the engine to ignite. When the engine fires, it exerts a constant 20 N horizontal thrust away from the wall. a. How far from the base of the wall does the rocket land?
Physics
1 answer:
pogonyaev3 years ago
5 0

Answer:

Explanation:

mass of rocket m = .5 kg

height of wall h = 40 m

initial horizontal velocity u = .5 m /s

horizontal acceleration = force / mass

= 20 / .5

a = 40 m /s²

Let rocket falls or covers 40 m vertically downwards in time t

h = 1/2 gt² , initial vertical velocity = 0

40 = 1/2 x 9.8 x t²

t = 2.8566 s

During this period it will cover horizontal distance with initial velocity of .5 m /s and acceleration a = 40m /s²

horizontal distance = ut + 1/2 at²

= .5 x 2.8566 + .5 x 40 x 2.8566²

= 1.4283 + 163.2033

= 164.63 m .

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