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

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- 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

thanks! im going to try it out

- 1018

@Michele_Laino this one

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- 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

ok!

- 1018

thank you so much!

- Michele_Laino

here we have a parabolic motion, the situation is like below:
|dw:1440003729324:dw|

- 1018

oh wait! sorry!

- 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

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

im sorry but it's this one. its actually the same but with different values sorry didnt see that

- Michele_Laino

ok! I see

- Michele_Laino

we have to compute the velocity when our projectile, passes at these points:
|dw:1440004078499:dw|

- Michele_Laino

namely at P1, and P2

- 1018

ok. im following so far

- 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

|dw:1440004431440:dw|

- 1018

what is m? and is g the gravitational acceleration?

- 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

yes, and that's all i got so far. at least it's correct. haha

- 1018

oh, i'll just see what you do. sorry im not familiar with using your formula, with the m one

- Michele_Laino

the equation for the vertical component of the velocity is:
\[\Large {v_y}\left( t \right) = {v_{0y}} - gt\]

- 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

so, after a substitution, we get:
\[\Large {v_y}\left( t \right) = {v_{0y}} - gt = \sqrt {2gh} - gt\]

- Michele_Laino

nevertheless we don't know the value of the maximum height h

- 1018

yes

- Michele_Laino

I think that we have insufficient data

- Michele_Laino

we need to know, at least, the measure of the maximum height

- 1018

sorry, but that is everything

- Michele_Laino

I think that many parabolic motions are possible with the horizontal velocity which you have provided

- 1018

someone answered this a while ago but i saw it late, maybe you can check out what he/she said

- Michele_Laino

I saw that answer

- Michele_Laino

I think that with your data, the motion it is not uniquely determined

- 1018

any other ideas how to do this?

- 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

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

im gonna move on from this, try to get some rest, study again tomorrow. haha

- Michele_Laino

yes! I'm interested, please tell me your solution :)

- 1018

ok, wait. i'll just make the answer wrong haha

- Michele_Laino

ok!

- 1018

here

##### 1 Attachment

- 1018

it's a printscreen of the solution

- Michele_Laino

yes! that is the solution, and I have failed, instead.
Sorry!!

- 1018

no no its ok. it's not like i figured it out too. haha. thank you so much anyway

- Michele_Laino

ok! :)

- 1018

hey, im gonna rest for awhile, i'll review this again tomorrow. haha. thank you again!

- Michele_Laino

:)

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