PHYSICS

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.

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  1. 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

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    posted by rohit

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