Golf Ball Projectile Motion Problem -- Solved

In summary, the conversation discusses finding the range using the given values of initial velocity (v0), angle (θ), and acceleration due to gravity (g). The correct equation for finding range is (2v0*cosθ * (v0sinθ)/g) and after correcting for units, the correct answer is 69.02. The discrepancy in the calculation is due to an error in using the calculator.
  • #1
Homework Statement
After a golf ball is hit it takes off with an initial speed of 26.6 m/s and at an angle of 36.5° with respect to the horizontal. The golf field is flat and horizontal. A) Neglecting air resistance how far will the golf ball fly? B) How high will the golf ball rise? C) How much time will the ball spend in the air? D) How far would the ball fly if the initial speed was doubled? E) How much time would the ball spend in the air in this second case? Thanks
Relevant Equations
Range = 2v0*cosθ * (v0sinθ)/g = (v0^2)/g *sin2θ
26.6m/s = v0
36.5º = θ
g = 9.81m/s^2

A) Find Range:
(v0^2)/g *sin2θ = (26.6^2)/9.81 * sin2(36.5)
= 89.99

I double-checked with the other Range equation (2v0*cosθ * (v0sinθ)/g) so I know I'm doing something wrong. Please help! Thank you

edit: A) Correct answer is 69.0
 
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  • #2
Undeterred247 said:
26.6m/s = v0
36.5º = θ
g = 9.81m/s^2

Find Range:
(v0^2)/g *sin2θ = (26.6^2)/9.81 * sin2(36.5)
= 89.99
(v0^2)/g *sin2θ = (26.6^2)/9.81 * sin2(36.5) is not 89.99 unless the units are not meters but something else. What you are doing wrong is between you and your calculator.
 
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  • #3
kuruman said:
(v0^2)/g *sin2θ = (26.6^2)/9.81 * sin2(36.5) is not 89.99 unless the units are not meters but something else. What you are doing wrong is between you and your calculator.
Oh my- Thank you!
 
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