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Chapter 4: Q. 46 (page 107)

A projectile鈥檚 horizontal range over level ground is $\frac{{v_{0}}^{2} \sin 2 胃}{g}$ . At what launch angle or angles will the projectile land at half of its maximum possible range?

Short Answer

Expert verified

The required angles are $15 掳 and 75 掳$ .

Step by step solution

01

Step 1. Given information

The formula to calculate the range $(R)$ of a projectile is given by

localid="1648260841963" $R = \frac{{v_{0}}^{2} \sin 2 胃}{g} . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

02

Step 2. Calculation for the condition of maximum range

The range of the projectile will be maximum when

$\sin 2 胃 = 1 . . . . . . . . . . . . . . . . . . . . . . (2)$

Solve equation (2) to calculate the required projection angle for which the range is maximum.

$\begin{array}{rcl} \sin 2 胃 & = & 1 \\ = & \sin 90 掳 \\ 2 胃 & = & 90 掳 \\ 胃 & = & 45 掳 \end{array}$

Since $\sin (180 掳 - 胃) = \sin 胃$ , the projectile will cover the maximum range for the projection angle being $胃$ or $(90 掳 - 胃)$ .

03

Step 3. Calculation for the condition of half the maximum range

The projectile will cover half the maximum range when

$\sin 2 胃 = \frac{1}{2} . . . . . . . . . . . . . . . . (3)$

Solve equation (3) to calculate one of the required projection angle.

$\begin{array}{rcl} \sin 2 胃 & = & \frac{1}{2} \\ = & \sin 30 掳 \\ 2 胃 & = & 30 掳 \\ 胃 & = & 15 掳 \end{array}$

The other value of the required projection angle is given by

$\begin{array}{rcl} (90 掳 - 胃) & = & 90 掳 - 15 掳 \\ = & 75 掳 \end{array}$

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