Motion - How do you find the time in the air of an object

[mncyapntsi]

Homework Statement

How do you find the time in the air of an object kicked horizontally from a slope of 42º, knowing the speed?

Relevant Equations

x = xo + vot + 1/2gt^2

Hi!
I am really stuck on the concept, wouldn't we not have enough variables?
I tried equation the equation for the y of the object to the slope of the hill, but I am missing several variables.
Any help is greatly appreciated :)
thanks!

 

[berkeman]

Homework Statement:: How do you find the time in the air of an object kicked horizontally from a slope of 42º, knowing the speed?
Relevant Equations:: x = xo + vot + 1/2gt^2

Hi!
I am really stuck on the concept, wouldn't we not have enough variables?
I tried equation the equation for the y of the object to the slope of the hill, but I am missing several variables.
Any help is greatly appreciated :)
thanks!

The question is a little confusing for me. Do you mean that the ball is kicked on flat ground at an upward initial angle of 42 degrees, and you need to figure out how long it takes for it to land a distance away? Or the ball is kicked up a sloping ramp of 42 degrees somehow?

If it is the first case, then figure out what the vertical component of that velocity is, and just solve for the vertical motion part to figure out how long it takes to land.

Also, you used x in your equation where we traditionally use y for the vertical in 2-D motion of a projectile...

 

[mncyapntsi]

The ground is at a 31º downward incline, but the ball is kicked at a 0º - purely horizontally. I hope that clears things up... What equation should I be using then...?

 

[PeroK]

The ground is at a 31º downward incline, but the ball is kicked at a 0º - purely horizontally. I hope that clears things up... What equation should I be using then...?

It's two-dimensional motion, right? So, that means calculating the $x$ and $y$ components.

Have you drawn a sketch? That might give you some ideas.

 

[berkeman]

The ground is at a 31º downward incline, but the ball is kicked at a 0º - purely horizontally. I hope that clears things up

Ah, that makes it a little harder, but still do-able.

What equation should I be using then...?

Using y for the vertical direction and x for the horizontal direction, and fixing the direction of g to be downward, you have the traditional two equations for projectile motion:
$$y(t) = y_0 + v_{y_0} t - \frac{1}{2}g t^2$$
$$x(t) = x_0 + v_{x_0} t$$
You have an equation for the slope of the downward incline (it is a straight line) and the equations for the x and y position as a function of time. Both lines start at your kick spot, and converge at some (x,y) point down the slope somewhere. You solve the equations simultaneously. Can you draw a sketch and start working with the equations?

 

[Steve4Physics]

Homework Statement:: How do you find the time in the air of an object kicked horizontally from a slope of 42º, knowing the speed?
Relevant Equations:: x = xo + vot + 1/2gt^2

Hi!
I am really stuck on the concept, wouldn't we not have enough variables?
I tried equation the equation for the y of the object to the slope of the hill, but I am missing several variables.
Any help is greatly appreciated :)
thanks!

So you really mean this:

The side of a hill is inclined at 42º (or perhaps 31º) to the horizontal.

A ball on the hill is kicked with an initial speed of v₀ so that the ball’s initial direction is horizontally outwards. The ball eventually hits the hill lower down. How long is the ball in the air?

The kicker falls due to attempting to kick a ball on such a steep hill, and later comes to rest, with only minor injuries, at the bottom.

 

[mncyapntsi]

It's two-dimensional motion, right? So, that means calculating the $x$ and $y$ components.

Have you drawn a sketch? That might give you some ideas.

Yes I drew a sketch and I have :
y(t) = yo+vo t-1/2 g t^2
y = -0.743145 x (with angle 42, sorry that was a typo)
x(t) = xo + vxo t
I really don't know where to go from here?

 

[mncyapntsi]

Ah, that makes it a little harder, but still do-able.

Using y for the vertical direction and x for the horizontal direction, and fixing the direction of g to be downward, you have the traditional two equations for projectile motion:
$$y(t) = y_0 + v_{y_0} t - \frac{1}{2}g t^2$$
$$x(t) = x_0 + v_{x_0} t$$
You have an equation for the slope of the downward incline (it is a straight line) and the equations for the x and y position as a function of time. Both lines start at your kick spot, and converge at some (x,y) point down the slope somewhere. You solve the equations simultaneously. Can you draw a sketch and start working with the equations?

Right, so I found
y = -0.743145 x (with angle 42, sorry that was a typo)
I'm not too sure how to solve these simultaneously

 

[PeroK]

Yes I drew a sketch and I have :
y(t) = yo+vo t-1/2 g t^2
y = -0.743145 x (with angle 42, sorry that was a typo)
x(t) = xo + vxo t
I really don't know where to go from here?

You know $y_0$, $v_{0_y}$, $x_0$ and $v_{0_x}$, right?

 

[mncyapntsi]

That's the thing -- I don't know y0 and x0. Could I pose them as being 0 and 0 ?
voy = 0
vox = cos(42)vo
Correct?

 

[berkeman]

That's the thing -- I don't know y0 and x0. Could I pose them as being 0 and 0 ?

Yes.

voy = 0
vox = cos(42)vo

$v_0$ is horizontal, so there is no cos() term, just the horizontal velocity.

 

[PeroK]

That's the thing -- I don't know y0 and x0. Could I pose them as being 0 and 0 ?
voy = 0
vox = cos(42)vo
Correct?

Your equation for the slope already assumes the top of the slope is the origin, right? In any case, you get to choose where you put the orgin. The top of the slope where the projectile starts seems simplest.

$v_{0_x}$ relates to the projectile. It has nothing to do with the slope.

 

[hmmm27]

Just to be sure, could you show us your sketch ?

 

[mncyapntsi]

Just to be sure, could you show us your sketch ?

Sure, here it is :

 

[mncyapntsi]

Your equation for the slope already assumes the top of the slope is the origin, right? In any case, you get to choose where you put the orgin. The top of the slope where the projectile starts seems simplest.

$v_{0_x}$ relates to the projectile. It has nothing to do with the slope.

Okay that clears things up a bit. I tried putting the y equation for the slope equal to the y equation for the projectile, but got a negative t :

Where did I go wrong...?

 

[PeroK]

What's the initial speed of the projectile?

 

[mncyapntsi]

What's the initial speed of the projectile?

66kmh

 

[PeroK]

66kmh

But, you've given an answer in seconds? What unit of time are you using - hours or seconds?

In terms of where you went wrong (apart from that), all I can see is a bunch of numbers. It's hard to see. The answer,obviously, can't be negative, so maybe you should sort that out as well. And see what you get next time.

Personally, I would use some algebra, as it makes things much clearer than a bunch of numbers.

 

[kuruman]

Okay that clears things up a bit. I tried putting the y equation for the slope equal to the y equation for the projectile, but got a negative t :
View attachment 289526
Where did I go wrong...?

$\cos(42^o)=\dfrac{x}{\sqrt{x^2+y^2}}~$, not what you have ($=\frac{x}{y}$).

 

[mncyapntsi]

But, you've given an answer in seconds? What unit of time are you using - hours or seconds?

In terms of where you went wrong (apart from that), all I can see is a bunch of numbers. It's hard to see. The answer,obviously, can't be negative, so maybe you should sort that out as well. And see what you get next time.

Personally, I would use some algebra, as it makes things much clearer than a bunch of numbers.

OH! i am so dumb of course. I forgot to convert Haha! Thanks!

 

[kuruman]

OH! i am so dumb of course. I forgot to convert Haha! Thanks!

If the conversion is the only correction you make, your answer will still be incorrect because the relation between $x$ and $y$ in your solution is incorrect as it stands.