Vertical Motion

Question

You shoot a basketball from an initial height of 7 feet and an initial velocity of 28 feet per second. The hoop is 10 feet tall. You goal is to find the amount of time in seconds that the shot is in the air. Which equation below best illustrates this scenario? h=in feet and t= in seconds

Answer 1

height is a function of time, h(t) = -16t^2 + vot + ho where vo = initial velocity = 28, and ho is initial height =7

If the shot misses the target, the hoop, then it stays in the air until it hits the floor at h=0. set the original equation = 0 and solve for t, by factoring, completing the square or the quadratic formula

set h(t) = 0 = -16t^2 + 28t +7 = 0

h'(t) = -32t + 28 = 0 gravity reduces velocity by 32 feet per second

t= 28/32 = 7/8 second for reaching maximum height = 0.875 seconds

to fall back to the floor would be very roughly twice that, with modification due to starting at 7, so close to 2 seconds, but under 2 seconds.

for the original quadratic equation h(t)

try the quadratic formula, h=-b/a + or - 1/a sqr(b^2-4ac) where a = -16, b=28 and c=7

t = 28/32 + or - (1/32)sqr(28^2+4(16)(7))

= 7/8 (maximum height) + or - (1/32)sqr (784+224)

= 7/8 + or - (1/32)sqr(1008)

= about 7/8 + or - (1/32)10(sqr10)

= about 7/8 + or - (10/32)(3.2)

= about 7/8 + or - 1, ignore the negative as it makes t<0

= about 1.87 seconds

= almost 2 seconds

that's if it misses the hoop or goes in and it's "nothing but net" causing almost no slowing as it continues onto the floor

h(1) = -16 +28 + 1 = 13 feet in the air

h(2) = -16(4) + 28(2) + 7 = -64 + 56 + 7 = -64 + 63 = -1 = 1 foot below the floor

slightly less than 2 seconds places the ball on the floor

a full 2 seconds, and it's already hit the floor

1<t<2

t is somewhere between 1 and 2, but very close to 2

But if the question meant how long is the ball in the air until it hits the hoop

work the problem the same way but instead of setting h(t) = 0, set h(t) = 10

and solve for t

After 1 second, it's still 13 feet in the air, and coming down.

to get to 10 feet, the height of the hoop would still make 1<t<2

but t would now be closer to 1

h(t) = -16t^2 + 28t + 7 = 10

-16t^2 +28T -3 = 0

t = 28/32 + (1/32)sqr(28^2 - 4(16)(3))

= 7/8 + (1/32)sqr(784-192)

=7/8 + (1/32)sqr592

= 7/8 + (1/32)24.4

=7/8 + 24.4/32

= about 7/8 + 6/8

= 13/8 = 1.37 seconds until the ball hits the hoop

Answer 2

The equation of motion is s(t) = 7 + 28t - 16t².

You want s(t)=10...but this equation will have 2 solutions: one as the ball goes up and one as it hits the rim.

Use the quadratic formula!