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Question 3 of 20 Let $h = \frac{v_0^2}{4.9}sin\theta cos\theta$ model the horizontal distance in meters traveled by a projectile. If the initial velocity is 44 meters/second, which equation would you use to find the angle needed to travel 150 meters? A. 197.55sin(2$\theta$) = 150 B. 395.10sin(2$\theta$) = 150 C. 150sin(2$\theta$) = 150 D. 8.98sin(2$\theta$) = 150

          Question 3 of 20
Let $h = \frac{v_0^2}{4.9}sin\theta cos\theta$ model the horizontal distance in meters traveled by a
projectile. If the initial velocity is 44 meters/second, which equation would
you use to find the angle needed to travel 150 meters?
A. 197.55sin(2$\theta$) = 150
B. 395.10sin(2$\theta$) = 150
C. 150sin(2$\theta$) = 150
D. 8.98sin(2$\theta$) = 150

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Question 3 of 20
Let h = (v0^2)/(4.9)sinθ cosθ model the horizontal distance in meters traveled by a
projectile. If the initial velocity is 44 meters/second, which equation would
you use to find the angle needed to travel 150 meters?
A. 197.55sin(2θ) = 150
B. 395.10sin(2θ) = 150
C. 150sin(2θ) = 150
D. 8.98sin(2θ) = 150

Added by Michael S.

Precalculus with Limits

Ron Larson 2nd Edition

Instant Answer

Solved by Expert Julie Silva

09/06/2022

Step 1

Step 1: Given that h=(v_(0)^(2))/(49)sinθcosθ models the horizontal distance in meters traveled by a projectile, where v0 = 44 m/s. Show more…

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Question 3 of 20 Let h=(v_(0)^(2))/(49)sin heta cos heta model the horizontal distance in meters traveled by a projectile. If the initial velocity is 44 meters/second, which equation would you use to find the angle needed to travel 150 meters? A. 197.55sin(2 heta )=150 B. 395.10sin(2 heta )=150 C. 150sin(2 heta )=150 D. 8.98sin(2 heta )=150 Question 3 of 20 Let h sin cose model the horizontal distance in meters traveled by a 49 projectile.If the initial velocity is 44 meters/second,which eguation would you use to find the angle needed to travel 150 meters? A.197.55sin2=150 B.395.10sin2=150 C.150sin2=150 D.8.98sin2=150

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Transcript

00:01 So in this problem, we're given this formula for h that represents the horizontal distance in meters that's traveled by a projectile.

00:07 So in part a, we're told the initial velocity is 20 meters per second.

00:10 So that would be our v -0 value.

00:13 So they want us to know what equation we would use to find the angle needed to travel 50 meters.

00:18 Well, 50 meters would be our h value because that's what h represents.

00:21 So in this case, all we're going to do is substitute 20 in place of v -0 and 50 in place of h.

00:27 So we're going to have 50 equals 20 squared divided by 32 times 2 times the sine of theta times the cosine of theta.

00:36 Now, i'm not sure how far you need to go ahead and simplify this.

00:40 So let's see if we can simplify this a little bit for us.

00:43 So we can start by multiplying both sides of our equation by 32.

00:47 So we're going to have 32 times 50.

00:49 That would equal to 1 ,600.

00:51 And then on the right hand side, we have 200 squared, which is 400.

00:55 So this is all equal to, and i'm going to still leave it like this, 2 sine of theta times the cosine of theta.

01:02 Well, 400 times 2 is equal to 800, or we could do a couple of things.

01:07 What i now can do is i can divide both sides of our equation by 400.

01:12 So we would have 4 equal to 2 sine theta cosine theta...

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