To determine the velocity of the projectile after 5 seconds, we can break down the initial velocity into its horizontal and vertical components.
Given: Initial velocity (v0) = 30 m/s Launch angle (θ) = 60°
The horizontal component of the velocity (v0x) can be found using the equation: v0x = v0 * cos(θ)
The vertical component of the velocity (v0y) can be found using the equation: v0y = v0 * sin(θ)
Using these equations, we can calculate the horizontal and vertical components of the velocity:
v0x = 30 m/s * cos(60°) v0x = 30 m/s * 0.5 v0x = 15 m/s
v0y = 30 m/s * sin(60°) v0y = 30 m/s * √3/2 v0y = 15√3 m/s
Now, we can find the final velocity (v) after 5 seconds by taking into account the acceleration due to gravity. Assuming no air resistance, the horizontal velocity remains constant, while the vertical velocity changes due to gravity.
The horizontal component of the velocity remains the same throughout the motion, so the horizontal velocity after 5 seconds is:
vx = v0x = 15 m/s
The vertical component of the velocity changes due to the acceleration due to gravity (g = 9.8 m/s²), and after 5 seconds, it can be calculated using the equation:
vy = v0y - g * t
where t is the time (5 seconds).
vy = 15√3 m/s - 9.8 m/s² * 5 s vy = 15√3 m/s - 49 m/s vy = -49 m/s + 15√3 m/s vy ≈ -5.08 m/s
The negative sign indicates that the velocity is directed downward.
Finally, we can calculate the magnitude and direction of the final velocity. The magnitude of the final velocity (v) can be found using the Pythagorean theorem:
v = √(vx² + vy²) v = √((15 m/s)² + (-5.08 m/s)²) v ≈ 16.05 m/s
The direction of the final velocity can be determined using trigonometry:
θ = atan(vy / vx) θ = atan((-5.08 m/s) / (15 m/s)) θ ≈ -18.87°
Note: The negative angle indicates that the direction is below the horizontal.
Therefore, after 5 seconds, the projectile has a magnitude of approximately 16.05 m/s and is directed at an angle of approximately -18.87° (below the horizontal).