Projectile Motion: Hitting a box with a softball

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SUMMARY

The discussion focuses on calculating the launch speed required for a softball to hit a box being hoisted at a constant speed of 10 m/s using a compressed-air cannon angled at 30 degrees. The key equation derived involves equating the vertical positions of both the softball and the box at the moment the softball has traveled 100 meters horizontally. The box's constant speed implies zero vertical acceleration, allowing the use of the equation for the box's position as a function of time without the influence of gravity on its vertical motion.

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Homework Statement


Students at an engineering contest use a compressed-air cannon to shoot a softball at a box being hoisted straight up at 10m/s by a crane. The cannon, tilted upward at a 30 degree angle, is 100m from the box and fires by remote control the instant the box leaves the ground. Students can control the launch speed of the softball by setting the air pressure. What launch speed should the students use to hit the box?

Homework Equations


The Attempt at a Solution


** This is an example from Physics for scientists and engineers - Knight and I don't understand how they got their position equation for the box.
So basically they say that you need to equate the vertical positions of the softball and the box at the instant that the ball has traveled 100m. so they then say:
(y1)s = (y1)b
((Voy)s)t - 1/2gt[tex]^{2}[/tex] = ((Voy)b)t
** so I don't understand why gravity would not act on the box as well? or are they just saying that since its moving at a constant speed, acceleration = 0 (so no gravity)..??
 
Last edited:
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Yeap you are right constant speed therefore acceleration is 0. No so long you are on the surface of Earth you will experience gravitational force unless say you are mageneto. The question stated that the box is moving at constant speed so the distance traveled would just be vt :biggrin:
 
Last edited:
thanks for clarification :)
 

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