If we neglect air resistance, then the range of a ball (or any projectile) shot at an angle θ with respect to the x axis and with an initial velocity v0, is given byR(θ)= (V0^2/g) sin(2θ)      for 0 ≤ θ ≤ pie/2 where g is the acceleration due to gravity (9.8 meters per second per second).For what value of θ is the maximum range attained? (Note that the answer is numerical, not symbolic.)

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Asked Oct 7, 2019
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If we neglect air resistance, then the range of a ball (or any projectile) shot at an angle θ with respect to the x axis and with an initial velocity v0, is given by

R(θ)= (V0^2/g) sin(2θ)      for 0 ≤ θ ≤ pie/2 


where g is the acceleration due to gravity (9.8 meters per second per second).

For what value of θ is the maximum range attained? (Note that the answer is numerical, not symbolic.)

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Expert Answer

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Step 1

For maximum range dR/dtheta=0

v0...

2.
Vo
R(O)=|
|sin(20)
2
dR
|(2)cos(20)
dR
=2
cos(20)
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2. Vo R(O)=| |sin(20) 2 dR |(2)cos(20) dR =2 cos(20)

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Math

Calculus