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

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

|dw:1444185191751:dw|

- anonymous

Where did you derive that from? Which kinematic equation?

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## More answers

- 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

it is a standard formula, which can be derived from the theory of falling bodies

- Michele_Laino

in our case we have: \(\theta=45\;degrees\)

- 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

I don't seem to understand that. We're only given three general equations. |dw:1444185562567:dw|

- anonymous

I'm just learning Physics by the way. :D

- 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

So I'm guessing I'll use the third formula?

- 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

|dw:1444185855561:dw|

- 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

Oh that's what I drew, so I guess I can keep up on what you're showing to me.

- Michele_Laino

yes! that's right!
the derivation of the formula for range \(r\) is long

- Michele_Laino

namely it is a long procedure

- anonymous

All right. I'll work on that. So what's the next step?

- anonymous

3.13 m/s?

- Michele_Laino

the maximum height reached by your grasshopper is the vertex of the parabola that I drew:
|dw:1444186256072:dw|

- Michele_Laino

correct! the initial velocity has a magnitude of \(3.13\) m/sec

- anonymous

if it's in the maximum height, then wouldn't \(V_0y=0\)

- Michele_Laino

correct! at maximum height the vertical velocity has magnitude equal to zero

- Michele_Laino

here is the formula for maximum height:
\[\Large {h_{\max }} = \frac{{v_0^2}}{{2g}}{\left( {\sin \theta } \right)^2}\]

- anonymous

0.25 m?

- 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

correct!

- anonymous

Oh okay. Thanks!

- Michele_Laino

:)

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