# Initial Velocity of a Rocket

## Homework Statement

I need to find the initial velocity of a rocket using the method a=$$\Delta$$v/$$\Delta$$t. Given a situation where $$\Delta$$t = 4 seconds. The force of gravity in this case is -10(m/s)

## Homework Equations

After I get $$\Delta$$v how do I turn that into the initial velocity?

## The Attempt at a Solution

a=$$\Delta$$ v/$$\Delta$$ t
-10=$$\Delta$$ v / 4 seconds
$$\Delta$$ v = -40 (m/s)

LowlyPion
Homework Helper
Welcome to PF.

What you found was the speed of an object dropped after 4 seconds ... or if you are talking about a rocket, the speed in accelerating from rest at 10 m/s2.

The acceleration of gravity while of interest doesn't say anything about the force the propellant delivers to the motion of a rocket.

This is just after launch so I see no need to worry about any force acting on the rocket while it's in the air other than gravity pulling it down.

drizzle
Gold Member
can you type the full question please, it sounds fuzzy like this

A rocket has been in the air for 4 seconds from the instant it was launched to when it hit the ground. The force of gravity can be rounded to 10(m/s). Using the method a= delta v / delta t, find the initial velocity for the rocket.

drizzle
Gold Member
you mean 10 (m/s^2)

wouldn't that motion of the rocket be a bow like motion [projectile motion]since it didn't mention that it was verticly launched!

if so work on the vertical axis, where the final velocity=0 and the intial one v [as you did], then try to find the distance from start point to hit point to get v horizontal which is constant [not accelerated] then add both to get the initial velocity [v_i=squareroot(v_v^2+v_h^2)]

LowlyPion
Homework Helper
This is just after launch so I see no need to worry about any force acting on the rocket while it's in the air other than gravity pulling it down.

OK. So it is not about the rocket at all except that it is a projectile after launch?

In which case the usual rules apply.

V = Vo - a*t = Vo - g*t = Vo - 10*t on the way up and then again on the way down.

With that in mind, then by symmetry it takes 2 seconds up, and 2 more down.

At 2 seconds to max height, then initial velocity is ...