anonymous
  • anonymous
PROJECTILE MOTION I don't know how to get started with this. A grasshopper jumps a horizontal distance of 1m from rest, with an initial velocity at a 45 degree angle with respect to the horizontal. Find (a) the initial speed of the grasshopper and (b) the maximum height reached.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Michele_Laino
  • Michele_Laino
here we have to apply the formula of the range \(r\), namely: \[r = \frac{{v_0^2}}{g}\sin \left( {2\theta } \right)\]
Michele_Laino
  • Michele_Laino
|dw:1444185191751:dw|
anonymous
  • anonymous
Where did you derive that from? Which kinematic equation?

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Michele_Laino
  • Michele_Laino
that formula expresses the range \(r\) of an object, which is launched with a velocity whose magnitude \(v_0\) and angle (with respect to the horizontal line) equal to \(\theta\)
Michele_Laino
  • Michele_Laino
it is a standard formula, which can be derived from the theory of falling bodies
Michele_Laino
  • Michele_Laino
in our case we have: \(\theta=45\;degrees\)
Michele_Laino
  • Michele_Laino
from my formula above, we get: \[{v_0} = \sqrt {\frac{{rg}}{{\sin \left( {2\theta } \right)}}} = \sqrt {\frac{{1 \cdot 9.81}}{{\sin \left( {2 \cdot 45} \right)}}} = ...m/\sec \]
anonymous
  • anonymous
I don't seem to understand that. We're only given three general equations. |dw:1444185562567:dw|
anonymous
  • anonymous
I'm just learning Physics by the way. :D
Michele_Laino
  • Michele_Laino
yes! we can derive my formula above starting from your formulas, nevertheless you have to keep in mind that the motion of the grasshopper is a composed motion
anonymous
  • anonymous
So I'm guessing I'll use the third formula?
Michele_Laino
  • Michele_Laino
better is if you follow my replies, furthermore the trajectory of the grasshopper is a parabola, or, in other words, we have a parabolic motion
Michele_Laino
  • Michele_Laino
|dw:1444185855561:dw|
anonymous
  • anonymous
And we can change \(V_0\) to \(V_0 sin \theta\)? Yeah, I'm sorry if I weren't paying attention to you. I just want to take it slow, I guess, since I'm just beginning to understand it. I just don't want that you're giving me a formula that I've never heard of.
anonymous
  • anonymous
Oh that's what I drew, so I guess I can keep up on what you're showing to me.
Michele_Laino
  • Michele_Laino
yes! that's right! the derivation of the formula for range \(r\) is long
Michele_Laino
  • Michele_Laino
namely it is a long procedure
anonymous
  • anonymous
All right. I'll work on that. So what's the next step?
anonymous
  • anonymous
3.13 m/s?
Michele_Laino
  • Michele_Laino
the maximum height reached by your grasshopper is the vertex of the parabola that I drew: |dw:1444186256072:dw|
Michele_Laino
  • Michele_Laino
correct! the initial velocity has a magnitude of \(3.13\) m/sec
anonymous
  • anonymous
if it's in the maximum height, then wouldn't \(V_0y=0\)
Michele_Laino
  • Michele_Laino
correct! at maximum height the vertical velocity has magnitude equal to zero
Michele_Laino
  • Michele_Laino
here is the formula for maximum height: \[\Large {h_{\max }} = \frac{{v_0^2}}{{2g}}{\left( {\sin \theta } \right)^2}\]
anonymous
  • anonymous
0.25 m?
Michele_Laino
  • Michele_Laino
substituting our data, we get: \[\Large {h_{\max }} = \frac{{v_0^2}}{{2g}}{\left( {\sin \theta } \right)^2} = \frac{{{{3.13}^2}}}{{2 \cdot 9.81}}{\left( {\sin 45} \right)^2}=...?\]
Michele_Laino
  • Michele_Laino
correct!
anonymous
  • anonymous
Oh okay. Thanks!
Michele_Laino
  • Michele_Laino
:)

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