blue7
Jan 2, 2009, 11:47 AM
Hey everyone! I have this physics problem I'm trying to solve, and I'm not sure if I did it right. Could someone tell me if its right?
The problem is:
A rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 75 m/s. It moves for 25s along its initial line of motion with an acceleration of 25 m/s^2. At this time, the engines fail, and the rocket proceeds to move as a free body.
a) What is the rocket's maximum altitude?
b) What is rocket's total time of flight?
c) What is the rocket's horizontal range?
My solution:
a) So I know that
a= 25
t=25s (until the engines fail)
v= 75 m/s
1st- I found my vyo= v x sin (x)
So in this case, its
vyo= 75 x sin (53)= 59.89m/s
2nd- Then, I figured since the engine failed, it stopped and started moving downward. Thus, at 25s, it would reach max height. Also, I didn't use g= -10 because its not free body at this point, and gravity isn't factoring into this, so I plugged in a= 25 for g.
So, I used d= vyot + 0.5gt^2 (or in this case, I'm using a for g)
d= (59.89)(25) + (0.5)(25)(625)
d= 1497.25 + 7812.5
d= 9309.75m
b) Now, this is the part I got confused about. I figured that since its now free body, I would use the same distance formula, but use -10 for g now. Then, I thought I would add this value to 25 seconds.
So, -9309.75m = 59.89t + .5 (-10t^2)
0= -5t^2 + 59.89t + 9309.75
0= 5t^2 - 59.89 - 9309.75
(quadratic equation) t= 49.55 s and then 49.55 + 25 is 74.55 seconds.
c) This part is easy, but again I'm not sure if I did a and b correctly.
vx= 75 x cos (53)
= 45.14 m/s
range= vx x t
range = 45.14 x 74.55
range= 3365.187
Thank you sooooo much if you reply to this!
The problem is:
A rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 75 m/s. It moves for 25s along its initial line of motion with an acceleration of 25 m/s^2. At this time, the engines fail, and the rocket proceeds to move as a free body.
a) What is the rocket's maximum altitude?
b) What is rocket's total time of flight?
c) What is the rocket's horizontal range?
My solution:
a) So I know that
a= 25
t=25s (until the engines fail)
v= 75 m/s
1st- I found my vyo= v x sin (x)
So in this case, its
vyo= 75 x sin (53)= 59.89m/s
2nd- Then, I figured since the engine failed, it stopped and started moving downward. Thus, at 25s, it would reach max height. Also, I didn't use g= -10 because its not free body at this point, and gravity isn't factoring into this, so I plugged in a= 25 for g.
So, I used d= vyot + 0.5gt^2 (or in this case, I'm using a for g)
d= (59.89)(25) + (0.5)(25)(625)
d= 1497.25 + 7812.5
d= 9309.75m
b) Now, this is the part I got confused about. I figured that since its now free body, I would use the same distance formula, but use -10 for g now. Then, I thought I would add this value to 25 seconds.
So, -9309.75m = 59.89t + .5 (-10t^2)
0= -5t^2 + 59.89t + 9309.75
0= 5t^2 - 59.89 - 9309.75
(quadratic equation) t= 49.55 s and then 49.55 + 25 is 74.55 seconds.
c) This part is easy, but again I'm not sure if I did a and b correctly.
vx= 75 x cos (53)
= 45.14 m/s
range= vx x t
range = 45.14 x 74.55
range= 3365.187
Thank you sooooo much if you reply to this!