2 Projectile motion Q's (1 Viewer)

[davidbarnes]

• “A rocket initially on the ground, aimed at an angle of 15.0 degrees from the horizontal was in the air 3.0 s before falling back to the ground.” What was its original speed?

-[FONT=&quot] [/FONT]Are we even given enough information to solve this?

• “A launcher sent a projectile horizontally from the top of a cliff overlooking the ocean at 20 ms^-1. It took 2.5 s before tit was seen to hit the water.”Determine a) the height of the cliff (and the launcher) b) the horizontal distance traveled by the projectile.

- Just can’t work this out, according to my calculations the projectile would either not leave the launcher, or would fall out of the launcher and blow the guy up lol.

 

[veloc1ty]

First question:

vy = uy + at
0 = uy - 9.8*3
uy = 29.4 m/s

uy = usinx
29.4 = usin15
u = 29.4 / sin15
u = 45.21 m/s



Second question:

(delta)y = uyt + 0.5at^2
(delta)y = 0*2.5 + 0.5*9.8*2.5^2
(delta)y = 30.625 m

(delta)x = uxt
(delta)x = 20*2.5
(delta)x = 50 m


Someone say something if I made any mistakes.

 

[davidbarnes]

Thanks for that Veloc1ty (nice name by the way).

On another note. I think the teahcer has got incorrect answers for the two below questions. Could somone please double check and tell me the snaers to the below. To me they seem simple and I get the same answer every time.

1. A space shuttle of mass of 7.5 x 10^4 kg is in orbit around the moon at a height of 100 km. What is the gravitational potential energy possessed by a body of mass 1.0 kg on its surface?
What is correct answer?

2. A body of mass 3.0 x 10^2 is situated on the surface of planet Z. It has a Gpe of -8.0 x 10^10 J. What is the mass of the planet?

 

[f(sex)]

shouldnt it be

First question:

vy = uy + at
0 = uy - 9.8*1.5
uy = 14.7m/s

uy = usinx
14.7 = usin15
u = 14.7 / sin15
u = 56.79 m/s

because the t is should only be 1.5 as v=0 at the maximum vertical displacement, which is half the time of the projectiles range...

 

[veloc1ty]

shouldnt it be

First question:

vy = uy + at
0 = uy - 9.8*1.5
uy = 14.7m/s

uy = usinx
14.7 = usin15
u = 14.7 / sin15
u = 56.79 m/s

because the t is should only be 1.5 as v=0 at the maximum vertical displacement, which is half the time of the projectiles range...

No, question says "in the air 3.0 s before falling back to the ground" which means it took 3 seconds to reach max height.

I'm not sure how to do those 2 new questions, how did you arrive at an answer?

 

[davidbarnes]

I'm not sure how to do those 2 new questions, how did you arrive at an answer?

Um, I used the Gravitational Potential Energy formula, don't know any other way to do them.

 

[veloc1ty]

Um, I used the Gravitational Potential Energy formula, don't know any other way to do them.

But you're not given the radius for either question...so how do you do it?

 

[kooltrainer]

velocity, ur wrong! its 1.5 second instead of 3 seconds.. and f(sex) was right.. it took 3 seconds to reach max then drop down (to the ground).. so 1.5 sec to reach max..
either way.. ur final answer for question 1 was wrong anyways.. u = 29.4 / sin15
"u" does not equal to 45.21 m/s ..