A basketball player jumped straight up to grab a rebound. If she was in the air for 0.80 second, how high did she jump, to the nearest hundredth of a meter?

2 Answers
Jul 13, 2017

#"height" = 0.78# #"m"# (ideally)

Explanation:

We're asked to find how long an object (person) is in the air, knowing that the total flight time was #0.80# #"s"#.

Idealized projectile motion is perfectly parabolic, which means the flight paths before and after it starts to fall are mirror images:

http://images.tutorvista.com

This also means that the time when it reaches its maximum height is exactly half of the time between when it stats and lands (which is #0.80# #"s"#). This means that the particle is at its maximum height at #color(red)(t = 0.40# #color(red)("s"#.

So this is the same situation if an object is dropped from an unknown height and lands #0.40# #"s"# after it is dropped.

With that being said, we can use the equation

#Deltay = v_(0y)t - 1/2g t^2#

where

  • #Deltay = ?#

  • #v_(0y) = 0# (equivalent to being dropped from rest at an unknown height)

  • #g = 9.81# #"m/s"#

  • #t = 0.40# #"s"#

Therefore, we have

#Deltay = (0)(0.40color(white)(l)"s") - 1/2(9.81color(white)(l)"m/s"^2)(0.40color(white)(l)"s")^2#

#= (4.905color(white)(l)"m/s"^2)(0.16color(white)(l)"s"^2)#

#= color(blue)(0.78# #color(blue)("m"#

to the nearest hundredth of a meter.

Jul 14, 2017

It is given that basket ball player jumped straight up in the air and landed back on ground after #0.8s#

Neglecting air friction we know that time taken for upwards flight is equal to time taken for downwards drop after reaching maximum height.

Also that velocity becomes zero at the point of maximum height.

As such the problem reduces to dropping of an object from a height #h# and time taken for the drop to ground #="Total time of flight"/2#.

#:.# Time taken for downward drop #=0.8/2=0.4s#

Suppose that she jumped with initial velocity#=u#
Also suppose that she jumped to a height #h#

Using following kinematic expression

#s=ut+1/2g t^2#

and inserting various quantities we get

#h=0xx0.4+1/2(9.81) 0.4^2#
#h=0.78m# rounded to nearest hundredth of a meter.