A basketball player, about to "dunk"the ball,jumps 76cm vertically. HOw much time does the player spend (a) in the top 15cm of this jump and (b) in the bottom 15cm? Does this help explain why such players seem to hang in the air at the tops of their jumps?
2007-01-19 16:59:56 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
to the last qn
my answer is yes,
u see while dunking(when they attain maximum height)their velocity is zero)hence the eye has more time to see the basketball player at that pose....hence players seem to hang in the air at the tops of their jumps!
initial velocity
v^2=u^2+2as;
0=u^2-2gs (negative acceleration)
u^2=2gs
u^2=2*9.8*76
hey i cant give u the answer jus like that...this is the method!if u dont get it message me ok!
to find the time a player spends below 15 cms and above can be calculated by using the formulae
s=ut +1/2at^2
2007-01-20 01:20:54 · answer #1 · answered by ash007 2 · 0⤊ 0⤋
you're missing a few more values
for example
the acceleration of gravity which works at 9.8 m /s^2
and possibly his velocity (m/s) as he jumps up
2007-01-19 17:05:50 · answer #2 · answered by arcticcroc 4 · 0⤊ 0⤋
straps as in on display????? if that's the case then move on menu button someplace round your display........ press it and seek for degauss pick it and your display will blink and the straps will move.....i suppose...... if it does not..... it demands maintenance.
2016-09-07 22:07:10 · answer #3 · answered by delsignore 4 · 0⤊ 0⤋