projectile on mars (Check my answers)?

Question

On the surface of Mars, a Martian launches a 2.5kg object off of a 75m tall cliff with initial velocity of 15 m/s an angle of 36.9 degrees above the horizontal. Assume g on Mars is 3.7

a) What are the horizontal and vertical components of the initial velocity?

I got Vox = 15 m/s and Voy = 9 m/s

b) What are the horizontal and vertical components of the object's acceleration once it leaves the Martians tentacle?

I got Ax = 0 m/s^2 and Ay = -3.7m/s^2

c) What is the vertical component of the displacement from when the object is released to when it hits the ground?

I got Y0 = 70 and Y= 0

d) How long is the object in the air?

I got 9.047s for the time

e) And find the horizontal displacement?

And I got 135.705m for the horizontal displacement.

Answer 1

a. Your Vox is wrong -- should be 12 m/sec. Take a look at your drawing to see why.
b. Correct.
c. This cannot be correct. The number must be at least 75 meters, because the projectile was that high to start with. What you need to do is to figure the height above the launch point from which, if the projectile were dropped from there, it would acquire a velocity of Voy when reaching the launch point. From this, get the time required to achieve that velocity, and use that number in part d.
d and e. I have not run these numbers; fix your earlier errors and re-calculate these.

Answer 2

I'd like to add to rhsaunders's response (which I pretty much agree with):
Careful choosing of the zero elevation will help in solving part d) of the problem. Thus, for this part, I would recommend setting the ground where the object ultimately crashes to be the zero elevation. Y0 = 70 m implies you set the zero elevation to be 5 m above the ground below the cliff, which would make calculations weird. Setting the zero elevation at the ground below the cliff gave me a time of 9.25 sec. Setting the zero elevation 5 m above the previous zero elevation yielded the 9.05 sec that resembles your time, so it seems you had the right idea, but just a wrong number.
Likewise, in part e), you seem to be doing the right math, but since you are plugging in the wrong numbers, the answer for that is also wrong. (I got somewhere in the neighborhood of 111 ft.)
Hope that made sense.

Answer 3

Vox = 15cos(36.9) = 11.995 m/s
V0y = 15sin(36.9) = 9.006 m/s

And, since you got the initial conditions wrong, all the rest of them don't have a prayer in hell of being correct.

HTH ☺


Doug

Answer 4

to stumble on the main suitable option top positioned vertical velocity equivalent to 0 and discover the time and discover its top at that element or at present use formulation h(max)=h(0)+v^2sin^2t/2g t is perspective of projection h(o) is preliminary top