Physics Kinematics help?
In basketball, hang is an illusion in which a player seems to weaken the gravitational acceleration while in midair. The illusion depends much on a skilled player's ability to rapidly shift the ball between hands during the flight, but might also be supported by the longer horizontal distance the player travels in the upper part of the jump than in the lower part. If a player jumps with an initial speed of v0 = 6.60 m/s at an angle of θ0 = 40.5°, what percent of the jump's range does the player spend in the upper half of the jump (between maximum height and half maximum height)?
The answer is 1/sqrt2, but I'm not sure how?
- billrussell42Lv 73 months ago
Vy = 6.6sin40.5 = 4.29 m/s (assuming that the angle is with the horizontal... you don't say)
Vx = 6.6cos40.5 = 5.02 m/s time to reach peak, t₁ = Vy/g = 0.438 sec
height of peak h = v²/2g = 4.29²/19.6 = 0.939 m
half that is 0.469 m
time to reach 0.469 m
d = ½at² + v₀t + d₀
0.469 = ½(–9.8)t² + 4.29t + 0
4.9t² – 4.29t + 0.469 = 0
solving for t
t₂ = 0.128 s (time spent from ground to half peak)
which means 0.438 – 0.128 = 0.310 s = t₃ (time spent from half peak to peak)
since horizontal speed is constant, we can just use time numbers, ie time spent above half peak and total time.
percent = 0.310/0.438 = 0.708 = 70.8%
PS 1/√2 = 0.707 is not a percent. Perhaps you mean 70.7% ?
probably could have done this in a general way, and left out the numbers and gotten the same number