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.

1. Michele_Laino

here we have to apply the formula of the range $$r$$, namely: $r = \frac{{v_0^2}}{g}\sin \left( {2\theta } \right)$

2. Michele_Laino

|dw:1444185191751:dw|

3. anonymous

Where did you derive that from? Which kinematic equation?

4. 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 $$\theta$$

5. Michele_Laino

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

6. Michele_Laino

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

7. Michele_Laino

from my formula above, we get: ${v_0} = \sqrt {\frac{{rg}}{{\sin \left( {2\theta } \right)}}} = \sqrt {\frac{{1 \cdot 9.81}}{{\sin \left( {2 \cdot 45} \right)}}} = ...m/\sec$

8. anonymous

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

9. anonymous

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

10. 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

11. anonymous

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

12. 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

13. Michele_Laino

|dw:1444185855561:dw|

14. 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.

15. anonymous

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

16. Michele_Laino

yes! that's right! the derivation of the formula for range $$r$$ is long

17. Michele_Laino

namely it is a long procedure

18. anonymous

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

19. anonymous

3.13 m/s?

20. Michele_Laino

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

21. Michele_Laino

correct! the initial velocity has a magnitude of $$3.13$$ m/sec

22. anonymous

if it's in the maximum height, then wouldn't $$V_0y=0$$

23. Michele_Laino

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

24. Michele_Laino

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

25. anonymous

0.25 m?

26. Michele_Laino

substituting our data, we get: $\Large {h_{\max }} = \frac{{v_0^2}}{{2g}}{\left( {\sin \theta } \right)^2} = \frac{{{{3.13}^2}}}{{2 \cdot 9.81}}{\left( {\sin 45} \right)^2}=...?$

27. Michele_Laino

correct!

28. anonymous

Oh okay. Thanks!

29. Michele_Laino

:)