# Thread: Rocket in the Air

1. ## Rocket in the Air

A rocket is fired straight up from the ground with an initial velocity of 900 feet per second.

(a) How long does it take the rocket to reach 4200 feet?

(b) When will the rocket hit the ground?

2. Originally Posted by emfn96
A rocket is fired straight up from the ground with an initial velocity of 900 feet per second.

(a) How long does it take the rocket to reach 4200 feet?
Solve.

$\frac{900ft}{1\times seconds} = \frac{4200ft}{x\times seconds}$

Wait...

You need to incorporate gravity...

3. ## No Gravity

Originally Posted by Anonymous1
Solve.

$\frac{900ft}{1\times seconds} = \frac{4200ft}{x\times seconds}$

Wait...

You need to incorporate gravity...
The question does not say to include gravity. Is any equation needed to solve this question?

4. Originally Posted by emfn96
The question does not say to include gravity. Is any equation needed to solve this question?
Well if there was no gravity the rocket would never come down...

5. ## Math Question Not Physics

Originally Posted by Anonymous1
Well if there was no gravity the rocket would never come down...
This question was taken from a precalculus book not a physics text or course. Can you give me more detail?

6. You need to use newton's third law of motion...

$(v_2)^2 =(v_1)^2 + 2as$

$s=$ max height height rocket ascends
$a=$ acceleration due to gravity (32.1522 ft/s it is negative as it is accting downward)
$v_1=$ initial vel(900 ft/s)
$v_2=$ final vel(0 m/s at top pos it doesnt have velocity)

Where $t=$ time in the air.

$(v2) = (v1) + at$

7. This is, in my opinion, a really bad problem. The equation Anonymous1 gives, as well as $s= -gt^2+ v_0t+ s_0$ will work for something thrown upward so that the only force is that of gravity. But the whole point of a rocket is that it continues firing as it goes up so that is NOT true.