# physics

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A basketball is thrown at 45° to the horizontal. The hoop is located 4.5m away horizontally at a height of 1m above the point of release. What is the required initial speed?

• physics -

Horizontal
Vcos45*t=4.5
Vertical:
Vsin45*t-4.9t^2=1

solve for t in the first equation (in terms of V), put that into the second, and solve for V

• physics -

how do u solve v?

• physics -

Put t=4.5/Vcos45 into the second equation, you only have V as the unknown. It is a quadratic, use the quadratic equation.

• physics -

Not good explanation

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