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1018
 one year ago
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
1018
 one year ago
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

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0thanks! im going to try it out

1018
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.1here we have a parabolic motion, the situation is like below: dw:1440003729324:dw

1018
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.1at 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
 one year ago
Best ResponseYou've already chosen the best response.0im sorry but it's this one. its actually the same but with different values sorry didnt see that

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to compute the velocity when our projectile, passes at these points: dw:1440004078499:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1namely at P1, and P2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here 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
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440004431440:dw

1018
 one year ago
Best ResponseYou've already chosen the best response.0what is m? and is g the gravitational acceleration?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1m 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
 one year ago
Best ResponseYou've already chosen the best response.0yes, and that's all i got so far. at least it's correct. haha

1018
 one year ago
Best ResponseYou've already chosen the best response.0oh, i'll just see what you do. sorry im not familiar with using your formula, with the m one

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the equation for the vertical component of the velocity is: \[\Large {v_y}\left( t \right) = {v_{0y}}  gt\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1where 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
 one year ago
Best ResponseYou've already chosen the best response.1so, after a substitution, we get: \[\Large {v_y}\left( t \right) = {v_{0y}}  gt = \sqrt {2gh}  gt\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1nevertheless we don't know the value of the maximum height h

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that we have insufficient data

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we need to know, at least, the measure of the maximum height

1018
 one year ago
Best ResponseYou've already chosen the best response.0sorry, but that is everything

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that many parabolic motions are possible with the horizontal velocity which you have provided

1018
 one year ago
Best ResponseYou've already chosen the best response.0someone answered this a while ago but i saw it late, maybe you can check out what he/she said

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I saw that answer

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that with your data, the motion it is not uniquely determined

1018
 one year ago
Best ResponseYou've already chosen the best response.0any other ideas how to do this?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I'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
 one year ago
Best ResponseYou've already chosen the best response.0that'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
 one year ago
Best ResponseYou've already chosen the best response.0im gonna move on from this, try to get some rest, study again tomorrow. haha

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! I'm interested, please tell me your solution :)

1018
 one year ago
Best ResponseYou've already chosen the best response.0ok, wait. i'll just make the answer wrong haha

1018
 one year ago
Best ResponseYou've already chosen the best response.0it's a printscreen of the solution

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! that is the solution, and I have failed, instead. Sorry!!

1018
 one year ago
Best ResponseYou've already chosen the best response.0no no its ok. it's not like i figured it out too. haha. thank you so much anyway

1018
 one year ago
Best ResponseYou've already chosen the best response.0hey, im gonna rest for awhile, i'll review this again tomorrow. haha. thank you again!
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