Asked by Alex
A basketball is thrown at 45° to the horizontal. The hoop is located 6.2 m away horizontally at a height of 1.1 m above the point of release. What is the required initial speed?
I know that I need to use d = vit + (0.5)at^2, but I then get stuck. Thanks!
I know that I need to use d = vit + (0.5)at^2, but I then get stuck. Thanks!
Answers
Answered by
Henry
Y^2 = Yo^2 + 2g*h = 0
Yo^2 - 19.6*1.1 = 0
Yo^2 = 21.56
Yo = 4.64 m/s.
Vo = Yo/Sin A = 4.64/sin45 = 6.57 m/s @
45o.
Yo^2 - 19.6*1.1 = 0
Yo^2 = 21.56
Yo = 4.64 m/s.
Vo = Yo/Sin A = 4.64/sin45 = 6.57 m/s @
45o.
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