I on my physics homework It's a 2-Dimensional Projectile Motion problem.

[esanoussi]

Homework Statement

A rocket is launched at an angle that is 53 degrees from the ground. It travels at an initial velocity of 25 m/s, and after a certain point, begins to fly at a constant speed of 25 m/s for 25 seconds. After 25 seconds, the rocket experiences failure and begins to crash toward the ground. What is the horizontal range of the rocket? How far has it traveled?

Givens:

Vi (Initial velocity): 25 m/s
t (Time): 25 seconds
ax (acceleration in the x direction): 0
ay (acceleration in the y direction once the rocket begins to crash): -9.8 m/s²
Vf (final velocity): ?
horizontal range, or $$\Delta$$x: ?

Homework Equations

Vf = Vi + at
$$\Delta$$x or $$\Delta$$y = Vi(x,y)t+(1/2)at²

I'm not too sure if they'll be necessary, but here are some basic trigonometric equations:

• 
  cos$$\theta$$ = adjacent side/hypotenuse
  sin$$\theta$$ = opposite side/hypotenuse
  tan$$\theta$$ = opposite side/adjacent side

The Attempt at a Solution

I couldn't find a solution, but with the givens I found, I made the following attempts and observations:

If the rocket launched at an initial velocity of 25m/s, reached a constant speed, and then fell from that distance, then it is safe to say that Vf = -25m/s.

Because Vi = 25m/s and Vf = -25m/s, it is safe to say that $$\Delta$$y = 0, because 25 + (-25) = 0.


y:
Vi(y) = 25sin53
$$\Delta$$y = 0
ay = -9.8m/s²
t = 25

x:
Vi(x) = 25cos53
$$\Delta$$x = ?
ax = 0
t = 25

$$\Delta$$x = 25cos53(25) + 0
$$\Delta$$x = (25cos53)(25)

Now, the question is: if I finish solving for $$\Delta$$x, will I get the answer to the question? Is that the horizontal range?

Can someone please try to help me by the end of the day? My homework is due tomorrow, and if I finish it correctly, I will earn 10 extra points toward my Physics Marking Period Exam (which I need badly)!

 

[JDHalfrack]

Just a point of clarification, are you sure "after a certain point, begins to fly at a constant speed of 25 m/s for 25 seconds" is correct? I think I have seen this question before (in a book I used to use), and I believe the rocket accelerates at 25 m/s² for 25 s. But, I may be wrong. It just seesm that "after a certain point" is not enough information to accurately solve this problem.

If I am right, the way you want to attack this problem is to split it up in three parts:
1. What happens while the rocket is accelerating at 53^o?
2. What happens between when the rocket's engines shut off and when it reaches it's highest point?
3. What happens on the way from its highest point to the ground?

Let me know if this helps.

 

[JDHalfrack]

Sorry for the double post, but here is the question from one of my old textbooks:

A rocket is launched at an angle of 53^o above the horizontal with an initial speed of 75 m/s, as shown in Figure 3-31 (not included). It moves for 25 s along its initial line of motion with an acceleration of 25 m/s². At this time its engines fail and the rocket proceeds to move as a free body.

a. What is the rocket's maximum altitude?
b. What is the rocket's total time of flight?
c. What is the rocket's horizontal range?

I guess there's enough differences between this question and the one in the OP. So, maybe they aren't the same.

 

[esanoussi]

I think you're right. The way the question is phrased in your book sounds a lot like the way my teacher said it. (Sorry; he dictates the questions to us and we're just supposed to jot down whatever we hear.)

In that case, I have no idea how to answer the problem. :( I realize that in this HW forum, we're supposed to solve the questions ourselves, which I have every intention of doing, but could you somehow steer me in the right direction, or let me know if I originally was going in the right direction?