Cricket jump impulse with 20g mass and 2.55m range

[odie5533]

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:
$$V_{y} = VSin30=\frac{1}{2}V$$
$$y = \frac{1}{2}Vt-\frac{1}{2}gt^2$$
Set to 0 to find when he lands:
$$0 = \frac{1}{2}Vt-\frac{1}{2}gt^2$$
$$t = \frac{V}{g}$$

$$V_{x} = VCos30$$
$$x(t) = (VCos30)t$$
Set to 2.55, the horizontal distance traveled when he lands:
$$2.55=\frac{V^2}{g}Cos30$$
$$V=\sqrt{\frac{2.55 g}{Cos30}} = 5.3736 m/s$$

So, the inital momentum is 0, since his velocity before the jump is 0.
$$J = mv_{2} - mv_{1} = mv_{2} - 0 = (20 g)(5.3736 m/s) - 0 = 107.47$$

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.

 

[nrqed]

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 question does not quite make sense (the last part). Are you sure it was not "what is the magnitude of the impulse and of the total force..." ?

The Attempt at a Solution

I started by finding the velocity needed for the jump:
$$V_{y} = VSin30=\frac{1}{2}V$$
$$y = \frac{1}{2}Vt-\frac{1}{2}gt^2$$
Set to 0 to find when he lands:
$$0 = \frac{1}{2}Vt-\frac{1}{2}gt^2$$
$$t = \frac{V}{g}$$

$$V_{x} = VCos30$$
$$x(t) = (VCos30)t$$
Set to 2.55, the horizontal distance traveled when he lands:
$$2.55=\frac{V^2}{g}Cos30$$
$$V=\sqrt{\frac{2.55 g}{Cos30}} = 5.3736 m/s$$

So, the inital momentum is 0, since his velocity before the jump is 0.
$$J = mv_{2} - mv_{1} = mv_{2} - 0 = (20 g)(5.3736 m/s) - 0 = 107.47$$

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.

I did not check all the numbers but the reasoning is good. If they ask for the magnitude of the total force this is where the delta time would come up. I am pretty sure this is what they were asking.

 

[odie5533]

I copied the question word for word. It took me a long time to figure out what it was asking, and I agree it makes more sense for it to be and. Ah well, my professor is notorious for typos and unanswerable questions, so this must just be another one. Thanks for the help.

 

[nrqed]

I copied the question word for word. It took me a long time to figure out what it was asking, and I agree it makes more sense for it to be and. Ah well, my professor is notorious for typos and unanswerable questions, so this must just be another one. Thanks for the help.

Ok. For the impulse itself no time is required.
I just noticed: write the units on your final answer. And if you want it in SI units, you better convert the mass to kilograms.