Angle elevation of gun=θ. ball is fired w/ velocity v ft/sec. Range of ball given by R=((v^2)sin2θ)/32 ft. what angle of elevation maximizes range?

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kjcdb8er's profile pic

kjcdb8er | Teacher | (Level 1) Associate Educator

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The answer to this question relies on maximizing the value given by R. To maximize a function, take it's derivative and set to zero. In this case, the variable is the angle of elevation, θ.

d/dθ (R) = d/dθ (v^2/32 * sin2θ)

To take this derivative, you need the chain rule since:

d/dx sin x = cos x

and,

d/dx (f( g(x) ) = f'(g(x)) * g'(x)

 

d/dθ (R) = d/dθ (v^2/32 * sin2θ) = v^2/32 * cos(2θ) * 2

0 = cos(2θ)  --> 2θ = pi/2, 3pi/4

θ = pi/4, 3pi/8

The correct answer is pi/4.

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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Given that the range to the angle (x) is:

R = v^2 *sin 2x / 32

To find the maximum values for the range, first we need to determine the critical points:

Let us differenctiate with respect to x:

R' = 2v^2 cos 2x / 32

    = v^2 cos2x/ 16 = 0

==> cos2x = 0

== . 2x =  pi/2 , 3pi/2

==> x =  pi/4, 3pi/4

Then the function will have 2 extreme values at x = pi/4 and x = 3pi/4

Now we need to determine which value is a maximum and which is a minimum.

We will Determine the second derivative.

R'' = -2v^2 sin2x / 16

      = -v^2 sin2x/ 8

When R'' > 0 ==> the function has a minimum

when R'' < 0 ==> the function has a maximum

R'' = -v^2 *sin2x / 8 < 0

==> sin2x must be positive

==> sin2x > 0

Then the angle should be in the second quadrant for the sin to be positive.

Then the maximum point is pi/4

and the minimum point is 3pi/4

Then the rangle is maximum when the angle = pi/4

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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The initial velocity of the ball is v ft/sec. The ball has the two components of velocity: horizontal velocity vcosx and  vertical velocity vsinx , where x is the angle of elevation.

Let the time taken by the ball be t to reach the ground. Then the the range of the ball is given by :

R = vtcosx....(1)

Also the height h of the ball at any time t is given by:

h = vtsinx-(1/2)gt^2, whwere g is the acceleration due to gravity. When the ball reaches the ground h = 0. Therefore  0 = vtsinx - (1/2)gt^2. Therefore when the ball reaches ground t = vtsinx = (1/2)gt^2. Or 2vt-gt^2 = 0, Or t(2vtsinx-gt) = 0.

 Or t = 2vsinx/g when h= 0. Substitituting in (1), we get:

R = 2v^2(cosxssinx)/g = v^2(sin2x)/g.

Therefore R(x)  is maximum when sin2x is maximum. We know that  maximum of sin2x is 1 , when 2x = 90 degrees. Or when x = 90/2 = 45 degrees.

Threfore the maximum horizontal displacement of the ball is (v^2)/g when angle of elevation of the initial speed of the ball is 45 degree. Since the value of acceleration due to gravity g is approximaly 32ft/sec^2, the maximum horizontal displacement of the ball is  v^2/32, when the angle of elevation  x = 45 degrees.

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