A projectile's horizontal range over level ground is v02sin⁡2θg\frac{$$v_0^2$$ \sin 2\theta}{g}. At what launch angle or angles will the projectile land at half of its maximum possible range?

Ch 04: Kinematics in Two Dimensions

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 04: Kinematics in Two Dimensions

문제 47

4장, 문제 47

A projectile's horizontal range over level ground is $v sub 0 squared sine 2 theta over g$ . At what launch angle or angles will the projectile land at half of its maximum possible range?

검증된 단계별 안내

Start by recalling the formula for the horizontal range of a projectile:

R = \(\frac{v_0^2 \sin(2\theta)}{g}\)

, where

v_0

is the initial velocity,

\(\theta\)

is the launch angle, and

g

is the acceleration due to gravity.

The maximum possible range occurs when

\(\sin\)(2\(\theta\)) = 1

, which happens at

2\(\theta\) = 90^\(\circ\)

\(\theta\) = 45^\(\circ\)

. Substituting this into the range formula gives the maximum range as

R_{\(\text{max}\)} = \(\frac{v_0^2}{g}\)

To find the angle(s) at which the projectile lands at half the maximum range, set the range equal to half of

R_{\(\text{max}\)}

\(\frac{v_0^2 \sin(2\theta)}{g}\) = \(\frac{1}{2}\) \(\cdot\) \(\frac{v_0^2}{g}\)

Simplify the equation by canceling

\(\frac{v_0^2}{g}\)

on both sides:

\(\sin\)(2\(\theta\)) = \(\frac{1}{2}\)

Solve for

2\(\theta\)

using the inverse sine function:

2\(\theta\) = \(\arcsin\)(\(\frac{1}{2}\))

. This gives two possible solutions for

2\(\theta\)

within one full rotation:

2\(\theta\) = 30^\(\circ\)

and

2\(\theta\) = 150^\(\circ\)

. Divide each by 2 to find the corresponding launch angles:

\(\theta\) = 15^\(\circ\)

and

\(\theta\) = 75^\(\circ\)

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and is subject to the force of gravity. It can be analyzed in two dimensions: horizontal and vertical. The horizontal motion is uniform, while the vertical motion is influenced by gravitational acceleration. Understanding the principles of projectile motion is essential for determining the range and trajectory of the projectile.

Range of a Projectile

The range of a projectile is the horizontal distance it travels before landing, which is influenced by its initial velocity, launch angle, and the acceleration due to gravity. The formula for the maximum range is given by R = v₀² sin 2θ/g, where v₀ is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. This relationship shows how the angle of launch affects the distance traveled.

Launch Angle

The launch angle is the angle at which a projectile is launched relative to the horizontal. It plays a critical role in determining the range and height of the projectile's trajectory. For maximum range, the optimal launch angle is 45 degrees. To find the angles that yield half of the maximum range, one must analyze the sine function in the range formula, which can yield multiple angles for a given range.