- #1
odie5533
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Homework Statement
A 20g cricket leaps through a 2.55 m horizontal distance. The initial velocity of the cricket makes an angle of 30 degress with the horizontal direction. If it takes the cricket 0.019s to leave the ground, what is the magnitude of the impulse of the total force on the cricket that gives it the speed needed for the jump?
The Attempt at a Solution
I started by finding the velocity needed for the jump:
[tex]V_{y} = VSin30=\frac{1}{2}V[/tex]
[tex]y = \frac{1}{2}Vt-\frac{1}{2}gt^2[/tex]
Set to 0 to find when he lands:
[tex]0 = \frac{1}{2}Vt-\frac{1}{2}gt^2[/tex]
[tex]t = \frac{V}{g}[/tex]
[tex]V_{x} = VCos30[/tex]
[tex]x(t) = (VCos30)t[/tex]
Set to 2.55, the horizontal distance traveled when he lands:
[tex]2.55=\frac{V^2}{g}Cos30[/tex]
[tex]V=\sqrt{\frac{2.55 g}{Cos30}} = 5.3736 m/s[/tex]
So, the inital momentum is 0, since his velocity before the jump is 0.
[tex]J = mv_{2} - mv_{1} = mv_{2} - 0 = (20 g)(5.3736 m/s) - 0 = 107.47[/tex]
I never liked impulse/momentum problems, and I was wondering if any of the above is correct. I just plugged numbers into the formulas and have no idea if I did it right =/ Also, I didn't use the time of 0.019s anywhere.