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

Answer:

θ = 39.8°

Explanation:

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

$tan\ \theta = \frac{v_y}{v_x}$

where,

θ = launch angle from the vertical = ?

vy = vertical component of velocity = 5 m/s

vx = horizontal component of velocity = 6 m/s

Therefore,

$tan\ \theta = \frac{5\ m/s}{6\ m/s}\\\\\theta = tan^{-1}(0.8333)$

θ = 39.8°