Question

A model rocket climbs 200 m in 4 seconds. If was moving 10 m/s to begin with what is its final velocity

Answer 1

Answer: 90 m/s

Step-by-step explanation:

For this situation we will use the following equations:

`y=V_(o)t+(1)/(2)at^(2)` (1)

`V=V_(o) + at` (2)

Where:

`y=200 m` is the height of the model rocket at 4 s

`V_(o)=10 m/s` is the initial velocity of the model rocket

`V` is the velocity of the rocket at 4 s

`t=4 s` is the time it takes to the model rocket to reach 200 m

`a` is the constant acceleration due gravity and the rocket's thrust

Firstly, from equation (1) we have to find
`a`:

`200 m=(10 m/s)(4 s)+(1)/(2)a(4 s)^(2)` (3)

`a=20 m/s^(2)` (4)

Now we have to substitute this value of
`a` in (2):

`V=10 m/s + (20 m/s^(2))(4 s)` (5)

Finally:

`V=90 m/s` This is the rocket's final velocity