Asked by applebottom
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?
Answers
Answered by
bobpursley
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
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
Answered by
applebottom
how do u solve v?
Answered by
bobpursley
Put t=4.5/Vcos45 into the second equation, you only have V as the unknown. It is a quadratic, use the quadratic equation.
Answered by
Anonymous
Not good explanation
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