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anonymous
 one year ago
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.
anonymous
 one year ago
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.

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0here we have to apply the formula of the range \(r\), namely: \[r = \frac{{v_0^2}}{g}\sin \left( {2\theta } \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444185191751:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where did you derive that from? Which kinematic equation?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0that 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
 one year ago
Best ResponseYou've already chosen the best response.0it is a standard formula, which can be derived from the theory of falling bodies

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0in our case we have: \(\theta=45\;degrees\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0from 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
 one year ago
Best ResponseYou've already chosen the best response.0I don't seem to understand that. We're only given three general equations. dw:1444185562567:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm just learning Physics by the way. :D

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0yes! 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
 one year ago
Best ResponseYou've already chosen the best response.0So I'm guessing I'll use the third formula?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0better 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444185855561:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh that's what I drew, so I guess I can keep up on what you're showing to me.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0yes! that's right! the derivation of the formula for range \(r\) is long

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0namely it is a long procedure

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0All right. I'll work on that. So what's the next step?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0the maximum height reached by your grasshopper is the vertex of the parabola that I drew: dw:1444186256072:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0correct! the initial velocity has a magnitude of \(3.13\) m/sec

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if it's in the maximum height, then wouldn't \(V_0y=0\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0correct! at maximum height the vertical velocity has magnitude equal to zero

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0here is the formula for maximum height: \[\Large {h_{\max }} = \frac{{v_0^2}}{{2g}}{\left( {\sin \theta } \right)^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0substituting our data, we get: \[\Large {h_{\max }} = \frac{{v_0^2}}{{2g}}{\left( {\sin \theta } \right)^2} = \frac{{{{3.13}^2}}}{{2 \cdot 9.81}}{\left( {\sin 45} \right)^2}=...?\]
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