A test rocket is fired vertically upward from a well. A catapultgives it an init

Question

A test rocket is fired vertically upward from a well. A catapultgives it an initial velocity of 79.5 m/s at ground level.Subsequently, its engines fire and it accelerates upward at 4.10until it reaches an altitude of 920 m. At that point its enginesfail, and the rocket goes into free fall, with an acceleration of-9.80

(a) How long is the rocket in motion above the ground?

Round your answer to the nearest tenth.

________ s

(b) What is its maximum altitude?

Round your answer to the nearest hundredth.

________ km

(c) What is its velocity just before it collides with theEarth?

Round your answer to the nearest m/s.

________ m/s

Explanation / Answer

You are given: v0 = 79.5 m/s a = 4.10 m/s2 h1 = 920 m g = -9.80 m/s2 ____________________ You need: t = ? hmax = ? vhit = ? We know that first part of motion is constant acceleration motionwith initial speed. We have equation: h1 = v0t + at2/2 You solve this for quadratic equation for t, and take the solutionthat makes sense (positive one). When you find the time of this part of motion, you can find thevelocity at point when engines fail using this equation: v1 = v0 + at After the engines fail, rocket is acting like body thrown upwards.Its initial velocity is v1, acceleration due to gravity is g, its current height ish1. You have these equations that describe the motion: v2 = v1 + gt2 (g = -9.80) At point of maximum height, v2 = 0, and t2 istime to reach this height after the engines fail. You plug thattime in formula for height: h2 = v1t2 +gt22/2, and calculate hmax =h1 + h2 To find the velocity of right before impact just use the equationfor free fall from height hmax.

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