anonymous
  • anonymous
An arrow is shot at 30(degrees) above the horizontal. It's velocity is 49 m/s, and it hits the target. a) What is the maximum height the arrow will attain? b) The target is at the height from which the arrow was shot. How far away is it? Show steps please. Test tommorow.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
you have an up component and an outward component that are independant of each other
amistre64
  • amistre64
|dw:1332719160694:dw|
amistre64
  • amistre64
the height of the arrow only cares about the 49sin(30) in quadratic equation

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amistre64
  • amistre64
since gravity pulls down on everything near the surface of the earth at say: 9.8 m/s/s we construct an equation h(t) = -4.9 t^2 +49sin(30)t + 0 and where the vertex of this parabola is defines your highest point
anonymous
  • anonymous
Don't I use the sine and cosine to calculate velocity?
amistre64
  • amistre64
velocity is already given as 49
amistre64
  • amistre64
you use the sin and cos to determine the distances obtained with regards to height and away from
anonymous
  • anonymous
I need to find V(f) that's what's messing me up. It's right before the arrow hits the target. I have a formula as y(max) = 0 + 49m/s(t) + 1/2(-9.8 m/s) t(squared) I need to have V(f) to use the formula Vf(squared) = V(i)Squared + 2 (a) (d)
amistre64
  • amistre64
since the vertex of the parabola is the middle of the parabola; and the target is at the same height as the shooting; just dbl the vertex to get total length
amistre64
  • amistre64
what does finding any velocity have to do with either quation: a) What is the maximum height the arrow will attain? b) The target is at the height from which the arrow was shot. How far away is it? ?
amistre64
  • amistre64
you need to find the time that the arrow is actually in the air
anonymous
  • anonymous
You can use your second formula to find the height, with V2 = 0 at the highest point. Then find the time to get there Double the time for the arrow to come back down and use the velocity in the x direction to find that distance.
anonymous
  • anonymous
If I use the triangle with 30(degrees) do I do sine 30 =x/49?
amistre64
  • amistre64
|dw:1332719642168:dw| where V is the vertex in my picture
amistre64
  • amistre64
yes, sin(30) = h/49
anonymous
  • anonymous
And that gives me y(max)? So do I double 49 to get total distance?
amistre64
  • amistre64
when we know the time taken to get to the high point; we double it and put it into the x=49cos(30) t
amistre64
  • amistre64
you double the time it takes to get to the high point; and use that in the distance component; 49cos(30) * t
anonymous
  • anonymous
I used the 49sine30, that equals 24.5. What did that just solve for? Velocity?
amistre64
  • amistre64
that would be the initial velocity in the upward direction of the arrows flight path
amistre64
  • amistre64
since height is an upward dimension; you need that to determine overall height
anonymous
  • anonymous
So, i'm slightly confused. Do I solve this equation in 2 parts? To solve for the Maximum height I require time, which equation do I use first?
amistre64
  • amistre64
\[h(t)=-\frac{1}{2}g\ t^2+V_ot +H_o\] \[v(t)=-gt+V_o\] when v(t) = 0 we are at the high point in this thing \[v(t)=0\ when\ t=\frac{V_o}{g}\]
amistre64
  • amistre64
use height first to establish a time that the arrow stays aloft
amistre64
  • amistre64
distance is determined by the cos part of it all since that is the direction that is going away from you 49cos(30) is your initial velocity in the outward direction; and this lasts for as long as the arrow is in the air; or rather when:\[t=2\frac{V_o}{g}\]that we found in the first part
anonymous
  • anonymous
So what do I plug into the V in the first equation h(t)? Are we solving for the h(t)? Does that resemble maximum height?
amistre64
  • amistre64
h(t) is what i named the height function with respect to time
amistre64
  • amistre64
the velocity of the arrow in the up direction is being slowed down by gravity until it sops; then it begins to fall back down
amistre64
  • amistre64
so, the time that it reaches a zero velocity is our highest point with regard to time
anonymous
  • anonymous
Let me know if you still need help on this.
anonymous
  • anonymous
If you're still confused: You said: I need to have V(f) to use the formula Vf(squared) = V(i)Squared + 2 (a) (d) In the x-direction, there is no acceleration (a), the velocity in the x direction that it has when it leaves is the same when it hits the target, so you can't use the above equation for that part of the problem. The only acceleration is caused by gravity, in the downward direction. So you can use that equation to find the height, because at that point, the velocity in the y direction is zero.

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