Asked by Jill Hefner
A 2.00 METER TALL BASKETBALL PLAYER ATTEMPTS A GOAL 10.00 METERS FROM THE BASKET(3.05 METERS HIGH). IF HE SHOOTS THE BALL AT A 45.0 DEGREE ANGLE, AT WHAT INITIAL SPEED MUST HE THROW THE BALL SO THAT IS GOES THROUGH THE HOOP WITHOUT STRIKING THE BACKBOARD?
Initial hi=2.0
final hf=3.05
horizontal distance=10
Here are the equations:
hf=Hi + Vsin45*t - 4.9 t^2
and
10= Vcos45*t
In the second equation, solve for t in terms of the numbers and V. Then plug that into the first equation for t, and solve for V. It will be a quadratic, use the quadratic equation.
Initial hi=2.0
final hf=3.05
horizontal distance=10
Here are the equations:
hf=Hi + Vsin45*t - 4.9 t^2
and
10= Vcos45*t
In the second equation, solve for t in terms of the numbers and V. Then plug that into the first equation for t, and solve for V. It will be a quadratic, use the quadratic equation.
Answers
Answered by
rohit
hf=Hi + Vsin45*t - 4.9 t^2
(v sin 45)t = 1.05 + 4.905 t^2 ___ (1)
(v cos 45)t = 10 ___ (2)
(2)/(1):
10 tan 45 = 1.05 + 4.905 t^2
Solve for t, t = 1.35 s
Substitute t in (2),
(v cos 45)1.35 = 10
v = 10.47 m s^-1
(v sin 45)t = 1.05 + 4.905 t^2 ___ (1)
(v cos 45)t = 10 ___ (2)
(2)/(1):
10 tan 45 = 1.05 + 4.905 t^2
Solve for t, t = 1.35 s
Substitute t in (2),
(v cos 45)1.35 = 10
v = 10.47 m s^-1
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.