A diver springs straight upward with a speed of 2.5 m/s from adiving platform th

Question

A diver springs straight upward with a speed of 2.5 m/s from adiving platform that is 2.5 m above the horizontal water surface.He eventually falls down and enters the water below. A) How long in seconds does it take the diver to hit the watersurface? B) What is the velocity, in the unit vector notation, of thediver at the instant of entering the water? A diver springs straight upward with a speed of 2.5 m/s from adiving platform that is 2.5 m above the horizontal water surface.He eventually falls down and enters the water below. A) How long in seconds does it take the diver to hit the watersurface? B) What is the velocity, in the unit vector notation, of thediver at the instant of entering the water?

Explanation / Answer

First find the distance to the height at whichv=0 vf2=vo2+2ad vf2=0 d= vo2/2a d= (2.5m/s)2/2(9.8m/s2) d= .32 m Now find time and velocity using the total distance. dt=2.5m +.32m = 2.82m d=vot +1/2at2 vo=0 t = 2d/a t = 2(2.82m)/(9.8m/s2)
t = .76s vf=vo +at vf=at vf=(9.8m/s2)(2.82m) vf=7.45m/s *EDIT - I forgot to include the total time in thissorry. So basically the time for the diver to get to the maximumheight is t=2d/a =2(.32)/9.8 =.26 t= .76+.26 = 1.02s *EDIT - I forgot to include the total time in thissorry. So basically the time for the diver to get to the maximumheight is t=2d/a =2(.32)/9.8 =.26 t= .76+.26 = 1.02s

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