Amodel rocket is launched vertically with an engine that is ignited at time t = 0. the engine provides an upward acceleration of 30 m/^2 for 2 seconds. upon reaching its maximum height, the rocket deploys a parachute, and then descends vertically toward the ground. a) determine the speed of the rocket after the 2s firing of the engine b) what maximum height will the rocket reach? c) at what time after t=0 will the maximum height be reached?

A. For the first questions, it asked for the speed of the rocket when it goes down. A stands for speed so you add the upward speed of 30 then slows down by 10 when it goes down so

a=30+10=40m/s2

B. This refer to speed of the rocket as it goes down so with the formula

V = u +at

V = 40*2 = 80m/s

Then solve for S

S = ut + ½^2

S = 1/2* 40 * 2*2 = 80m

C. At t=0, the rocket is at rest.

answer is : muchas/tipicas

have a nice day!

~recklesschicken

a .4y4

hope this !