An athlete on a trampoline leaps straight up into the air with an initial speed of 9.0 m/.s. Find (a) the maximum height reached by the athlete relative to the trampoline (b) the speed of the athlete when she is halfway up to her maximum height.
2007-10-28 09:50:57 · 1 answers · asked by Mika 2 in Science & Mathematics ➔ Physics
It's easier to solve both parts by dropping an athlete from the topmost point and ask: (a) How far does he fall if is final speed is -9.0 m/s? (b) Half way to the bottom, how fast was he going?
(a) It takes time t = (-9.0 m/s)/(-g) for an acceleration -g to produce a speed of -9.0 m/s. During this time, the athlete falls a distance of (1/2) g * t^2. Use the calculated value of t from above.
(b) How long does it take to fall half that distance? Solve for t_h using:
-9.0/2 = (1/2)(-g)(t_h)^2
t_h = sqrt(9/g)
During the time t_h, the athlete accelerates from 0 to v_h = -g*t_h. Compute v_h and you're done.
The advantage to starting from the top, in this case, is that is the point in time and space that the athlete is motionless, so no adjustments are needed in the equations for initial velocity.
A drawing helps too, but I can't do that here.
2007-10-29 03:23:57 · answer #1 · answered by husoski 7 · 1⤊ 0⤋