A student finds that it takes .20s for a ball to pass throughphotogates placed 3

Question

A student finds that it takes .20s for a ball to pass throughphotogates placed 30 cm apart on a level ramp. Then end of the rampis 92 cm above the floor.
a.Where should a coin be placed so that the ball strikes itdirectly on impact?
b. suppose now that the same ball, released from the same ramp(92cm high) struck a coin placed 25 cm from the ramp. What was theball's horizontal velocity?
c. How long did it take for the ball to pass through thephotogates?

for part a. I know you do .3m/.2s to find the velocity at 1.5 m/s..but then I am not sure if I use that as Vyi.. I did use it as Vyithough and after I converted to meters I got .059s and found thedistance to be .0885m. Is that correct?

Then for part b and c I am not sure how to calculate for timebecause im not sure what formula to use. please help!

Explanation / Answer

a) v = 1.5 m/s (you are right!) this is vxi h = 0.92 m, vyi = 0 h = gt2/2, so t = (2h/g) = 0.433 s distance d = vt = 0.65 m b) d = 0.25 m, t = 0.433 s, v = d/t = 0.577 m/s c) 0.3/0.577 = 0.52 s

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