Calculating Time and Distance for Basketball Throw

[runner1738]

nelect: air friction
your teacher tosses a basketball. the ball gets through the hoop(lucky shot). the man stand 2.746 m and the throws the ball at an angle of 53 degrees with a velocity of 16 m/s, the hoop is 3.048 m.
note: the distance he is from the hook is not given.

the time it takes for the ball to reach its max height is 1.3, but how do you figure out long it takes to get to the hoop, then the next question is what is l being the length from the hoop

My issues: is the ball hits the hoop before it reaches the end of the parobola so how do you calculate that with the information given.

 

[runner1738]

is there anyway i can load the picture? maybe that will help?

 

[runner1738]

if the difference in height is .302 and you are given vi and theta can you calculate how long it takes for the ball to reach that height? then subtract that value from the total time the ball would have been in the air for? i know it take 1.3 s to reach max h

 

[Leong]

$$s=ut+\frac{1}{2}at^2$$

 

[Astronuc]

The parabolic trajectory can be described as a parametic set of equations.

One should be able to describe x = x(t) and y = y(t), also v_x = constant (because no air resistance) and v_y(t) = v_yo-gt.

So the time to get to the hoop is simply = d/v_x, where d is distance between hoop and man, or it can be determined from the time it takes to get from launch point (man's hand) elevation to peak and then down again to the hoop elevation. The peak is where v_y(t) = 0.

 

[runner1738]

the equation i used was rf=ri+vit+1/2gt^2 and solved for t then subtracted t from 2.6 being the total time of the parobola, but how can i find the length x between the man and the hoop, now that i know the time it takes to get there?

 

[Astronuc]

So the time to get to the hoop is simply t = d/v_x, where d is distance between hoop and man, or conversely, d = v_x * t (no air resistance means no deceleration in the x-direction).