Expert: Steve Holleran - 3/22/2007

Question
When slam-dunking, a basketball player seems to hang in the air at the height of his jump.  the height h(t), in feet above the ground, of the basketball player at time t, in seconds since the start of a jump, is given by:

h(t) = -16t(squared) + 16Tt

where T is the total time in seconds that it takes to complete the jump.  For a jump that takes 1 second to complete, how much of this time does the basketball player spend at the top 25% of the trajectory?

I keep getting to 1.19 seconds, but am unsure about the 25% versus 75%.  HELP :(

Answer
Hi Heidi,

Well, I'm not really sure that I fully understand whats going on here, but let me give it a try.

Since T = 1, our equation is h(t) = -16t^2 + 16t.
Also, if she completes the jump in 1 second, then she reaches the peak of her jump in 1/2 second. (You can also get this by finding the vertex of the parabola at t=-b/2a,
which would be t = -16/-32=1/2).

Okay, then, h(2)= -16(1/2)^2 + 16(1/2) = 4 feet.
So the top 25% of her trajectory (?) would be 3 feet or higher.  Now set h=3 and solve for t:

   3 = -16t^2 + 16t  -------> 16t^2 - 16t +3 = 0
and either factor this or use the quad. formula and get
t=1/4, and t=3/4.  So then she spends 3/4 - 1/4 = 1/2 second at the top 25%.

That's what it would seem be to me.  I hope I interpreted it correctly.  I'm  not really sure where the 1.19 sec is coming from.

Steve Holleran