Assume: A 78g basketball is launched at an angle of 42.3 degrees and a distance of 13.4m from the basketball goal. The ball is released at the same height (ten feet) as the basketball goal's height.
A basketball player tries to make a long jump-shot as described above.
The acceleration of gravity is 9.8m/s^2. What speed must the player give the ball? Answer in units of m/s.
Please help!
A basketball player tries to make a long jump-shot as described above.
The acceleration of gravity is 9.8m/s^2. What speed must the player give the ball? Answer in units of m/s.
Please help!
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use the range equation:
R = v0^2 sin(2 theta)/g
R=range = 13.4 m
v0=initial velocity
theta = launch angle = 42.3 deg
g=9.8m/s/s
solve for v0:
v0=Sqrt[g R/sin(2*theta)] = Sqrt[9.8m/s/s*13.4m/sin(84.6)] = 11.5m/s
R = v0^2 sin(2 theta)/g
R=range = 13.4 m
v0=initial velocity
theta = launch angle = 42.3 deg
g=9.8m/s/s
solve for v0:
v0=Sqrt[g R/sin(2*theta)] = Sqrt[9.8m/s/s*13.4m/sin(84.6)] = 11.5m/s