Question

What two angles of elevation will enable a projectile to reach a target 15 km downrange on the same level as the gun if the​ projectile's initial speed is 421 ​m/sec?

Answer 1

Angle of projection: 28 degrees or 62 degrees

Explanation:

A projectile consists of two independent motions:

• A uniform motion along the horizontal direction
• A uniformly accelerated motion along the vertical direction

From the equations of motion, it is possible to derive an expression for the range of a projectile, which is:

$d=\frac{u^2 sin(2\theta)}{g}$

where:

u is the initial speed

$\theta$ is the angle of projection

$g=9.8 m/s^2$ is the acceleration due to gravity

For the projectile in this problem, we have:

d = 15 km = 15,000 m is the range

$u=421 m/s$ is the initial speed

Solving the equation for $\theta$, we find the possible angle:

$sin (2\theta) = \frac{dg}{u^2}=\frac{(15,000)(9.8)}{(421)^2}=0.829$

Which gives two values of the angle:

$2\theta = 56^{\circ} \rightarrow \theta=28^{\circ}\\2\theta=124^{\circ} \rightarrow \theta=62^{\circ}$

Learn more about projectile:

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