A projectile is fired from ground level with a speed of 150 m/s at an angle 30.° above the horizontal on an airless planet where g = 10.0 m/s2. What is the horizontal component of its velocity after 4.0 s? Express your answer to two significant figures Group of answer choices

129.9 m/s = 130 m/s to 2 s.f

Explanation:

For projectile motion, the initial velocity combined with the angle of launch is used to obtain the initial horizontal and vertical components of the velocity.

u = initial velocity of the projectile = 150 m/s

uₓ = u cos θ = 150 cos 30° = 129.9 m/s

uᵧ = u sin θ = 150 sin 30° = 75.0 m/s

In the motion of a projectile, the motion can literally be separated into vertical and horizontal components.

The vertical component has to do with the acceleration due to gravity (acting downwards) on the projectile and the vertical component of the velocity (which changes all through the motion of the projectile because of the force of gravity manifested in the form of acceleration due to gravity).

But there is no force acting in the horizontal direction, hence, no acceleration in the horizontal direction for projectile motion. This directly translates to a constant velocity in the horizontal direction all through the flight of the projectile.

Hence, the horizontal component of the velocity of the projectile at t = 4 s is the same as the horizontal component of the initial velocity.

Horizontal component of the Velocity at t = 4s is equal to uₓ = u cos θ = 150 cos 30° = 129.9 m/s = 130 m/s

129.9 m/s.

Explanation:

Vo = 150 m/s

θ = 30°

g = 10 m/s^2

Horizontal component, Vx = Vo cos θ

= 150 × cos 30

= 150 × 0.866

= 129.9 m/s.

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