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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

Expert verified

The required angles are15°and75°.

Step by step solution

01

Step 1. Given information

The formula to calculate the rangeRof a projectile is given by

localid="1648260841963" R=v02sin2θg.........................(1)

02

Step 2. Calculation for the condition of maximum range

The range of the projectile will be maximum when

sin2θ=1......................(2)

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

sin2θ=1=sin90°2θ=90°θ=45°

Since sin180°-θ=sinθ, the projectile will cover the maximum range for the projection angle beingθor90°-θ.

03

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

The projectile will cover half the maximum range when

sin2θ=12................(3)

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

sin2θ=12=sin30°2θ=30°θ=15°

The other value of the required projection angle is given by

90°-θ=90°-15°=75°

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