Physics help please I know that I am supposed to use constant accl formula but that is it ?

Question

A basketball player grabbing a rebound jumps 70.6 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 10.5 cm of this jump and (b) in the bottom 10.5 cm? Do your results explain why such players seem to hang in the air at the top of a jump?

Any help would be appreciated

Answer 1

yes that's it,   s = ut +- 1/2 g t^2   and v^2 = u^2 +- 2g s

g is deceleration of -10m/s/s going up until he stops , then +10m/s/s as he falls back.

Answer 2

Weight pulls objects to Earth. When you jump you are providing a force against graivity, but as gravitational accelaration is constant it provides a constant force (weight) that pulls us to the Earth.

Consider the apex of the jump 0.706m from here he falls 0.105m. This takes some time. You'll also note at the apex of the jump he is not traveling up anymore. Hence we can consider just the accelration due to gravity.

One model is S = ut + 1/2 at^2

Ignore the ut because the inital velocity is zero. Hence;

0.105 = 1/2 at^2

0.21=at^2

a in this case is g (9.81 meter per second per second) so;

0.21 / 9.81 = t^2

t^2 = 0.0214...

t = 0.1463 s

Double this for when he was coming up 10.5cm

and the total time he spent 'hanging air' is: 0.29 of a second.

Compare that result to the full time to drop 70.6 doubled :)

Answer 3

v^2=u^2-(2.g.s), for s= .706m, v=0. Hence: 0=u^2-(2*10*.706), Thus: u=3.757 m/s.  

Now, s=ut-1/2(gt^2). Take the .105m distance. Then,

   .105=3.757t - 1/2(108t^2). This gives t= .7223sec and .0291 sec.

This signifies the times after start when the player's at the 10.5 cm mark.

So, he spends .0291*2=.0582 secs below 10.5 cm, and .7223-.0291=.6932 secs above 10.5 cms.

All the best. Pl get back in case of queries or suggestions.