Need some physics help regarding projectile motion?

Question

Your teacher tosses a basketball and goes through(lucky shot).

Initial velocity is 18 m/s, the angle is 62 degrees. The hoop is 3.048 m tall and the guy is 2.256 m tall. I already found my max height by using the formula and the max height is 1.62 sec.

I need to know how long it takes to reach the hoop and horizontal length and I'm stuck right now.

Thanks to all who help.

Answer 1

The equation of motion for the vertical displacement is
y(t)=2.256+18*sin(62)*t-.5*g*t^2
when y(t)=3.048 and the ball is descending, swoosh.
3.048=2.256+18*sin(62)*t-.5*g*t^2
solve for the larger t
t=3.19 seconds

that's the time to the hoop

the distance is
x(3.19)=18*cos(62)*3.19
27.0 m
That is a lucky shot
NBA and NCAA courts are 94 ft x 50 ft
27 m is 88.6 ft, that's like other foul line to swoosh.

j

Answer 2

heigth of hoop relative to tosser = 3.048 - 2.256 = 0.792m
v[x] = 18 cos(62) = 8.45 m/s
v[y] = 18 sin(62) = 15.893 m/s
0 = 15.893 - 980t
t = 0.01622 s
h = 15.893(0.01622) - (1/2)(980)(0.01622)^2 = 0.12887 m
Since the heigth of the hoop is higher than the maximum heigth of the toss, the data of this problem is suspect.

Answer 3

If the piece separates gently (and not with an explosion as in case of a fire cracker rocket ) then it will fly side by side as no external force has acted on the stone or the piece. So the correct answer is option 1. In case of an explosion midway it will depend on the point of separation. The small piece can go in any random direction (not necessarily horizontal) and follow different parabolic path.