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## Homework Statement

Ok, for my Mechanical engineering class we have been assigned to make a 2ft ^3 5kg device that will launch a tennis ball 20-40 ft. My group decided to use a design more or less like a crossbow, so I was hoping someone would look over my calculations and make sure they're right. I'm doing a rough calculation to make sure the bungee cords we use are strong enough to reach the max distance.

Here's what I've got.

Assume a tennis ball ways about 57g, will be accelerated over 1.75 ft at 45 degrees from horizontal, on a vertical flat surface

## Homework Equations

F=ma

V

_{f}= [itex]\sqrt{2a \Delta X}[/itex]

a

^{2}+b

^{2}=c

^{2}

-Vx and Vy make a right triangle with V

V

_{y}+ g*t= -V

_{y}

since it's height is the same at launch and impact it's velocity should be the opposite of it's initial.

2V

_{y}= -g*t

## The Attempt at a Solution

sorry, the itex started breaking on me, and I couldn't figure out where...

Vf=sqrt(2(F/56.7g) * 1.75ft)= sqrt(Vx^2 + Vy^2)

Vx=Vy

t = 40ft/Vx

2(F/56.7g)*1.75ft=2Vy^2=-gt

2(F/56.7g)*1.75ft=2Vy^2=-g*40ft/Vy

Vy^3=-(-9.81m/s^2)*40ft /2

Vy^3=(9.81m/s^2)*20ft

Vy=(3.911m/s)

2(F/56.7g)*1.75ft=2(3.911m/s)^2

F=1.626 newtons, and this would need to be the average force over the time of acceleration, so the initial force would be 2x because at the end position it would be 0 force so (2x +0)/2= x, thus 3.2 N ?

Thanks in advance.

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