Question

A projectile is launched from the ground with an initial horizontal velocity of 6.0m/s [right] and an initial vertical velocity of 5.0/s [up]. At what angle from vertical is it launched?

Answer 1

θ = 39.8°

Step-by-step explanation:

Her, we will use the trigonometric ratio of the tangent to find the launch angle of the projectile from the vertical:

`tan\ \theta = (v_y)/(v_x)`

where,

θ = launch angle from the vertical = ?

vy = vertical component of velocity = 5 m/s

vx = horizontal component of velocity = 6 m/s

Therefore,

`tan\ \theta = (5\ m/s)/(6\ m/s)\\\\\theta = tan^(-1)(0.8333)`

θ = 39.8°