A projectile was fired with initial velocity of 40 m/s and at angle 30⁰ above the horizontal. Its initial vertical velocity is 20 m/s.
In a projectile motion, at a time instant t, the velocity v can be divided into its horizontal and vertical components: vx and vy. Let the projectile is fired at angle θ above the horizontal, then:
vx = v cos θ
vy = v sin θ
Where:
v = velocity at time instant t
Parameters given:
v = 40 m/s
θ = 30⁰
Hence, the vertical velocity is:
vy = v sin 30⁰
= 40 . 1/2 = 20 m/s
Complete question:
Determine the magnitude of the initial vertical velocity of the projectile, vy, when the magnitude of its initial velocity, v, was 40. meters per second and the angle is 30⁰ above the horizontal.
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