A bottle rocket is launched 68 degrees above the horizontal at a velocity of 5.49m/s.

The initial velocity is 5.49m/s. The acceleration is -9.81m/s. The angle is 68 degrees.

Hi, jessie102!

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If you post what you think might be the answer, and how you arrived at the answer, you'll be more likely to have someone knowledgeable come along to verify whether you're correct or not and if not correct, help you to learn how to come up with the correct answer yourself.

Thanks!

Sorry, I missed these threads :o

Ok, start by having a sketch, that is very helpful in any problem involving motion.

1. Calculate the components of the bottle rocket. This is done using trigonometry.

The vertical component is given by 5.49sin(68) = 5.07 m/s
The horizontal component is given by 5.49cos(68) = 2.06 m/s

2. Now, taking only the vertical component, it is as if you are throwing it up then it falls back.

Use

v - final velocity
u - initial velocity
a - acceleration
s - displacement

You know the initial velocity (5.07), the acceleration (-9.81 [negative since it is in the opposite direction of your initial velocity]) and the final velocity is zero, which is when it attains its maximum height.

Solve for s, the displacement, or distance at which it reaches the maximum height.

3. For the distance one, it is a little more complicated, for you need to two formulae: and

FIrst, find the time of flight. This is obtained using the vertical component only.

Use .

You know the final velocity, v, at which the ball reaches its maximum height, the initial velocity u and the acceleration. Find the time t.

Now, that time is for half the total cistance, and that is only the upward journey. Multiply by two to get the total time of flight.

Since the horizontal component is constant throughout, you needed that time of flight. Now, you can find the distance:

Use .

You have u, the initial horizontal velocity, the time of flight, the acceleration is zero along the horizontal plane. Solve for s and you're done! :)

I hope this helped! :)