Expert: Richard J. Raridon - 5/7/2007

Question
Kelvin plays basketball and is known for his dunks. He also loves math and decided to study the mathematics of dunking. He learned in Algebra class that the height h in feet of an object t seconds after it is projected upward with an initial velocity of v ft/sec is found using the formula:

h=vt-16t^2

In Physics class Kelvin measured that when he jumps he leaves the floor with an initial velocity of 13 feet per second. The basket rim is 10 feet off the ground and he must get teh tip of his hand at least 6 inches above the rim to be able to dunk. He can reach a height of 99 inches before jumping. Determine how long Kelvin has to execute his dunk. In other words, for how much time is the tip of his hand at least 6 inches above the rim?
Extra: With his initial velocity of 13 feet per secound, how many inches off teh ground does Kelvin jump?


Answer
In this case you need to use the more general expression,
h = h0 +vt -16t^2
where h0 = 99 inches = 8.25 ft
so you have 10.5 = 8.25 +13t -16t^2
Solving for t gives you 0.25 and 0.5625s so he is above the rim for 0.3125s
For the second part,
v = v0 -gt
so 0 = 13 -32t and t = 0.40625 at the max. height
h = 8.25 +13(0.40625) -16(0.40625)^2 = 10.9 ft = 130.7 inches