1018
  • 1018
The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 19 m/s. (a) What is the magnitude of the velocity of the projectile 1.7 s before it achieves its maximum height? (b) What is the magnitude of the velocity of the projectile 1.7 s after it achieves its maximum height? If we take x = 0 and y = 0 to be at the point of maximum height and positive x to be in the direction of the velocity there, what are the (c) x coordinate and (d) y coordinate of the projectile 1.7 s before it reaches its maximum height and the (e) x coordinate and (f) y coordinate
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
IrishBoy123
  • IrishBoy123
the first sentence tells you the projectile's horizontal velocity, which you will assume to be constant in this problem and that point you also know that its vertical velocity = 0 and that, more generally, acceleration in the vertical direction is -g. armed with that information, all you need are the equations of motion to solve https://gyazo.com/002724abd9cb7eb055122e07be79d344
1018
  • 1018
thanks! im going to try it out
1018
  • 1018
@Michele_Laino this one

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

1018
  • 1018
i hate to ask, but can you also solve it? its worth alot of points on my homework and i hate to waste it
Michele_Laino
  • Michele_Laino
ok!
1018
  • 1018
thank you so much!
Michele_Laino
  • Michele_Laino
here we have a parabolic motion, the situation is like below: |dw:1440003729324:dw|
1018
  • 1018
oh wait! sorry!
1018
  • 1018
The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. (a) What is the magnitude of the velocity of the projectile 1.2 s before it achieves its maximum height? (b) What is the magnitude of the velocity of the projectile 1.2 s after it achieves its maximum height? If we take x = 0 and y = 0 to be at the point of maximum height and positive x to be in the direction of the velocity there, what are the (c) x coordinate and (d) y coordinate of the projectile 1.2 s before it reaches its maximum height and the (e) x coordinate and (f) y coordinate 1.2 s after it reaches its maximum height?
Michele_Laino
  • Michele_Laino
at that point the kinetic energy is: \[\Large KE = \frac{1}{2}m{v^2}\] where v is the velocity in my drawing, which is a horizontal vector
1018
  • 1018
im sorry but it's this one. its actually the same but with different values sorry didnt see that
Michele_Laino
  • Michele_Laino
ok! I see
Michele_Laino
  • Michele_Laino
we have to compute the velocity when our projectile, passes at these points: |dw:1440004078499:dw|
Michele_Laino
  • Michele_Laino
namely at P1, and P2
1018
  • 1018
ok. im following so far
Michele_Laino
  • Michele_Laino
here we have to keep in mind that the initial vertical velocity Vy is given by the subsequent formula: \[\Large \frac{1}{2}mv_y^2 = mgh\] where h is the maximum height: |dw:1440004390763:dw|
Michele_Laino
  • Michele_Laino
|dw:1440004431440:dw|
1018
  • 1018
what is m? and is g the gravitational acceleration?
Michele_Laino
  • Michele_Laino
m is the mass of our projectile, and g is gravity In a parabolic motion the horizontal velocity, doesn't change, so along all motion, the horizontal velocity is: \[\Large {v_x} = 14m/\sec \]
1018
  • 1018
yes, and that's all i got so far. at least it's correct. haha
1018
  • 1018
oh, i'll just see what you do. sorry im not familiar with using your formula, with the m one
Michele_Laino
  • Michele_Laino
the equation for the vertical component of the velocity is: \[\Large {v_y}\left( t \right) = {v_{0y}} - gt\]
Michele_Laino
  • Michele_Laino
where g is gravity and V0y is: \[\Large {v_{0y}} = \sqrt {2gh} \] which comes from my formula above: \[\Large \frac{1}{2}mv_y^2 = mgh\]
Michele_Laino
  • Michele_Laino
so, after a substitution, we get: \[\Large {v_y}\left( t \right) = {v_{0y}} - gt = \sqrt {2gh} - gt\]
Michele_Laino
  • Michele_Laino
nevertheless we don't know the value of the maximum height h
1018
  • 1018
yes
Michele_Laino
  • Michele_Laino
I think that we have insufficient data
Michele_Laino
  • Michele_Laino
we need to know, at least, the measure of the maximum height
1018
  • 1018
sorry, but that is everything
Michele_Laino
  • Michele_Laino
I think that many parabolic motions are possible with the horizontal velocity which you have provided
1018
  • 1018
someone answered this a while ago but i saw it late, maybe you can check out what he/she said
Michele_Laino
  • Michele_Laino
I saw that answer
Michele_Laino
  • Michele_Laino
I think that with your data, the motion it is not uniquely determined
1018
  • 1018
any other ideas how to do this?
Michele_Laino
  • Michele_Laino
I'm sorry, no my idea is that we need to know the maximum height or another parameter, for example the angle between the initial velocity and the horizontal line |dw:1440006104583:dw|
1018
  • 1018
that's ok. thank you so much anyway. hey if you're interested, i could printscreen the solution. it shows the solution when after you answered incorrectly. haha
1018
  • 1018
im gonna move on from this, try to get some rest, study again tomorrow. haha
Michele_Laino
  • Michele_Laino
yes! I'm interested, please tell me your solution :)
1018
  • 1018
ok, wait. i'll just make the answer wrong haha
Michele_Laino
  • Michele_Laino
ok!
1018
  • 1018
here
1 Attachment
1018
  • 1018
it's a printscreen of the solution
Michele_Laino
  • Michele_Laino
yes! that is the solution, and I have failed, instead. Sorry!!
1018
  • 1018
no no its ok. it's not like i figured it out too. haha. thank you so much anyway
Michele_Laino
  • Michele_Laino
ok! :)
1018
  • 1018
hey, im gonna rest for awhile, i'll review this again tomorrow. haha. thank you again!
Michele_Laino
  • Michele_Laino
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.