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A toy rocket is launched with an initial velocity of 10.0 m/s in the horizontal direction from the roof of a 38.0-m-tall building. The rocket's engine produces a horizontal acceleration of (1.60 m/s^3)t, in the same direction as the initial velocity, but in the vertical direction the acceleration is g, downward. Air resistance can be neglected.

Required:
What horizontal distance does the rocket travel before reaching the ground?

1 Answer

7 votes

Answer:

Step-by-step explanation:

Consider downward displacement first .

downward displacement h = 38 m .

downward acceleration = g

downward initial velocity u = 0

h = ut + 1/2 g t ²

38 = 0 + 9.8 t ²

t = 1.97 s

During this period , there will be horizontal displacement with initial velocity u = 10 m /s and acceleration a = 1.6 m /s²

s = ut + 1/2 a t²

= 10 x 1.97 + .5 x 1.6 x 1.97²

= 19.7 + 3.10

= 22.8 m

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