Finding height and range of a projectile

[!!!]

Homework Statement

A test rocket is launched by accelerating it along a 200.0-m incline at 1.25 m/s² starting from rest at point A. The incline rises at 35.0° above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance is ignored). Find:
a) the maximum height above the ground that the rocket reaches
b) the greatest horizontal range of the rocket beyond point A.

Homework Equations

v_x = v₀cosα₀
v_y = v₀sinα₀
Basically, all the projectile motion formulas.

The Attempt at a Solution

Well, I don't even know how to start because I'm horrible at identifying which to find first. I assume I should find the initial velocity (v₀) of the ball, then time, then find its max. height for a). But for part b), I have no clue. And I'm confused at whether to plug in 1.25 m/s² or 9.8 m/s² as the acceleration. How do I get to the answer of 124 m for a) and 280 m for b) Can someone walk me through this, please?

 

[Kurdt]

Lets take it step by step. What is the speed of the rocket when it reaches the end of the ramp? You know the acceleration and the distance and that it started from rest (i.e. its initial velocity was zero).

 

[!!!]

Won't the speed of the rocket be zero since at the highest point? Because the final y-velocity is zero, and the final x-velocity is

, so it's zero since it's the same as the initial x-velocity.

 

[Kurdt]

Its launched from the ramp. The question states it is accelerated along it until the instant the rocket leaves it. What is the speed when it launches from the ramp?

 

[!!!]

I can't think of a kind of formula that relates distance, acceleration and speed. .___.

 

[!!!]

But how do I know what's the initial velocity if I don't know the time? D:

 

[Kurdt]

Have a look at the following list.

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

 

[!!!]

Wait, am I supposed to find the x or y-component of <br /> v^2 = v_0^2 + 2 a \Delta x<br />, and do I plug in the 1.25 acceleration or gravity? =/

 

[Kurdt]

The acceleration is along the slope so you need not worry about the components.

 

[modulus]

Well, before beginning, I’ll tell you that it was great to see that you’ve posted a projectile motion question, because I just love projectile motion!

Anyways, let me clear things up first. The final velocity of the rocket for it’s journey on the inclined plane will act as the initial velocity for the rocket in it’s trajectory (and yes, then you will resolute that velocity).

The reason you are not being able to find the range correctly is because you are not considering the distance the rocket covered while it was on the inclined plane. It will be equal to 200m*cosѲ (this will work in a similar manner as the time when you found how much height the rocket gained while on the inclined plane, when finding the maximum height).

But, that’s not where the complications end. You must also consider the distance covered while coming down to the ground from the height it gained at the beginning (due to the inclined plane). I deduced a formula for finding out the total range from the point when it leaves the inclination to the point when it touches the ground again: ucosѲ {[usinѲ/g] + [sqrt (2h/g)]}.

Note that here ‘h’ refers to the maximum height which you deduced in the beginning; the one which included the height gained by the inclined plane. Also, you’ll have to add the distance covered on the inclination with the value you get from this.

I deduced this formula simply by adding the time taken by the rocket to cover the distance to the point of it’s maximum height (when it’s vertical velocity is zero) and the time taken to cover the distance from that point to the ground. After that I multiplied it by the initial velocity on the x-axis.

I used this formula, and, after a lot of approximation (I repeat: a lot of approximation) I came with the answer 264m. I’m pretty sure that if you’ll use the exact values and correct calculations, you’ll come up with 280m.


P.S.- I know the language of my post gets a bit confusing at times, but, if you just quick-read through it a few times, I’m sure you’ll get it!