Asked by Billy bob
A golf ball leaves a tee at 60n/s and strikes the ground 200m away. At what two angles with the horizontal could it have begin its flight? Find the timed of flight and maximum attitude in each case.
Answers
Answered by
R_scott
let's use m/s instead of n/s
the range equation is ... R = u^2 * sin(2θ) / g
... u is the launch velocity, and θ is the launch angle
solving for θ ... sin(2θ) = R * g / u^2
... there are two solutions
find the two possible vertical velocities ... u * sin(θ)
the flight times (t) are ... t = [2 * u * sin(θ)] / g
the max altitudes are ... 1/2 * g * t^2
the range equation is ... R = u^2 * sin(2θ) / g
... u is the launch velocity, and θ is the launch angle
solving for θ ... sin(2θ) = R * g / u^2
... there are two solutions
find the two possible vertical velocities ... u * sin(θ)
the flight times (t) are ... t = [2 * u * sin(θ)] / g
the max altitudes are ... 1/2 * g * t^2
Answered by
Billy bob
What's the answer? It's for college
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