Question

Determine el valor del ángulo de tiro de un proyectil que tarda en impactar en 8s y que tuvo un alcance de 0,3km.

Answer 1

Answer:

$46.3^{\circ}$

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform motion along the horizontal direction (constant velocity)

- A uniformly accelerated motion along the vertical direction (constant acceleration)

The time of flight of a projectile is given by

$t=\frac{2u_y}{g}$

where

$u_y$ is the initial vertical velocity

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

Here the time of flight is

t = 8 s

So, the initial vertical velocity is:

$u_y=\frac{gt}{2}=\frac{(9.8)(8)}{2}=39.2 m/s$

At the same time, the horizontal distance covered by the projectile is given by

$d=u_x t$

where

$u_x$ is the horizontal velocity, which is constant

t is the time of flight

Here we know that the range of the projectile is 0.3 km, so

d = 0.3 km = 300 m

So the horizontal velocity is

$u_x=\frac{d}{t}=\frac{300}{8}=37.5 m/s$

Therefore, we can find the angle of projection of the projectile by using:

$\theta=tan^{-1}(\frac{u_y}{u_x})=tan^{-1}(\frac{39.2}{37.5})=46.3^{\circ}$