Math Problem Statement

The range R and the maximum height H of a projectile fired at an inclination thetaθ to the horizontal with initial speed v 0v0 are given by the formulas​ below, where g almost equals 32.2g≈32.2 feet per second per second is the acceleration due to gravity. Complete parts A and B.Upper R equals StartFraction 2 v Subscript 0 Superscript 2 Baseline sine theta cosine theta Over g EndFractionR= 2v20sinθcosθ g     Upper H equals StartFraction v Subscript 0 Superscript 2 Baseline sine squared theta Over 2 g EndFractionH= v20sin2θ 2gA. Find the range R if the projectile is fired at an angle of 45 degrees45° to the horizontal with an initial speed of 160160 feet per second.

Solution

To find the range RR of the projectile when it is fired at an angle of θ=45\theta = 45^\circ with an initial speed of v0=160v_0 = 160 feet per second, we use the formula for range:

R=2v02sinθcosθgR = \frac{2v_0^2 \sin\theta \cos\theta}{g}

Step 1: Plug in the given values

  • v0=160v_0 = 160 feet per second
  • θ=45\theta = 45^\circ
  • g=32.2g = 32.2 feet per second squared

Step 2: Simplify the trigonometric functions

Since θ=45\theta = 45^\circ, we know: sin45=cos45=22\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}

Step 3: Substitute into the formula

R=2(160)2sin45cos4532.2R = \frac{2(160)^2 \sin 45^\circ \cos 45^\circ}{32.2}

R=2(160)2(22)(22)32.2R = \frac{2(160)^2 \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{2}}{2}\right)}{32.2}

R=2(160)2(24)32.2R = \frac{2(160)^2 \left(\frac{2}{4}\right)}{32.2}

R=160232.2R = \frac{160^2}{32.2}

Step 4: Compute the result

Now, calculate 1602160^2 and then divide by 32.232.2.

1602=25600160^2 = 25600

R=2560032.2795.03 feetR = \frac{25600}{32.2} \approx 795.03 \text{ feet}

So, the range RR is approximately 795.03 feet.

Would you like the details of the computation, or do you have any other questions?

Here are some related questions to consider:

  1. How does the range change if the angle is increased to 6060^\circ?
  2. What would the maximum height be for the same projectile at 4545^\circ?
  3. How does the initial speed affect the range of the projectile?
  4. What happens to the range if the angle is decreased to 3030^\circ?
  5. How would you calculate the time of flight for the projectile?

Tip: Remember that the angle 4545^\circ gives the maximum range for any given initial speed, assuming no air resistance.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Trigonometry
Kinematics

Formulas

Range of projectile formula: R = (2 * v0^2 * sinθ * cosθ) / g

Theorems

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Suitable Grade Level

High School