Chapter 4: Q. 46 (page 107)

A projectile’s horizontal range over level ground is $\frac{{v_{0}}^{2} \sin 2 θ}{g}$ . 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 $15 ° and 75 °$ .

Step by step solution

Step 1. Given information

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

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

Step 2. Calculation for the condition of maximum range

The range of the projectile will be maximum when

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

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

$\begin{array}{rcl} \sin 2 θ & = & 1 \\ = & \sin 90 ° \\ 2 θ & = & 90 ° \\ θ & = & 45 ° \end{array}$

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

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

The projectile will cover half the maximum range when

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

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

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

The other value of the required projection angle is given by

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

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

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation for the condition of maximum range

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

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A projectile’s horizontal range over level ground is v02sin2θg. At what launch angle or angles will the projectile land at half of its maximum possible range?

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculation for the condition of maximum range

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

One App. One Place for Learning.

Most popular questions from this chapter

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