Question

A toy rocket is launched vertically from ground level (y = 0.00 m), at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 72 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance. The speed of the rocket upon impact on the ground is closest to ...?

Show all the steps please..

Answer 1

you should use this equation Vfin^2= Vin^2 +2*g*y

Where

Vin= 30m/s

y=72 m then you can get what the Vfin is.

Answer 2

Hi,
let's assume gravitational acceleration ; g = 10 m/s

Basically there are 2 stages in this motion which we have to substitute linear motion formulas in order to solve the problem
1. motion due to deceleration ( until the rocket reach it's maximum height after engine burnout )
2. motion due to gravitational acceleration (motion after rocket reach it maximum height )

Let's start with the 1st stage
initial velocity = 30 m/s
displacement from the ground = 72m

let's find the distance that rocket can travel until it reaches it's maximum height.
V^2 = U^2 + 2AS


V = final velocity
U = initial velocity
A = gravitational deceleration
S = displacement

0 = 900 - 20S
S = 900/20
= 45 m

So rocket can travel another 45 m until it reaches it's maximum height
Th total displacement when the rocket is at it's maximum height = 72 +45
= 117 m

So, at the beginning of the second stage the displacement of the rocket is 117 m
now lets find the velocity that rocket will gain when it reaches the ground

displacement = 117 m
initial velocity = o m/s

V^2 = U^2 + 2AS
= 0^2 + (2*10*117)
= 2340
v = 48.37 m/s

Hey, this is the longer version method of solving this but if u r interest about the short method i could post that one too. I used this way because u have ask for explanations

Answer 3

u r welcome