Understanding the Problem
We have a projectile (the 5 kg object) launched horizontally from a height. We need to find various aspects of its motion, such as:
* Time of flight: How long it takes to hit the ground.
* Horizontal range: How far it travels horizontally before hitting the ground.
* Final velocity: Its velocity (speed and direction) just before impact.
Key Concepts
* Projectile Motion: The motion of an object launched into the air, subject only to gravity.
* Independence of Motion: The horizontal and vertical components of projectile motion are independent. This means:
* The horizontal velocity remains constant (ignoring air resistance).
* The vertical velocity is affected only by gravity.
Calculations
1. Vertical Motion
* Initial vertical velocity (viy): 0 m/s (since the object is launched horizontally)
* Acceleration due to gravity (g): -9.8 m/s² (negative since it acts downwards)
* Vertical displacement (Δy): -275 m (negative since it moves downwards)
We can use the following kinematic equation to find the time of flight (t):
Δy = viyt + (1/2)gt²
-275 = (0)t + (1/2)(-9.8)t²
t² = 56.12
t ≈ 7.49 s
2. Horizontal Motion
* Horizontal velocity (vix): 45 m/s (remains constant)
* Time of flight (t): 7.49 s (from previous calculation)
To find the horizontal range (Δx), we use:
Δx = vixt
Δx = (45 m/s)(7.49 s)
Δx ≈ 337.05 m
3. Final Velocity
* Horizontal velocity (vfx): 45 m/s (remains constant)
* Vertical velocity (vfy): We can find this using:
vfy = viy + gt
vfy = 0 + (-9.8 m/s²)(7.49 s)
vfy ≈ -73.4 m/s (negative indicates downward direction)
To find the magnitude of the final velocity (vf):
vf = √(vfx² + vfy²)
vf = √(45² + (-73.4)²)
vf ≈ 86.5 m/s
To find the angle (θ) of the final velocity:
θ = tan⁻¹(vfy / vfx)
θ = tan⁻¹(-73.4 / 45)
θ ≈ -58.1° (measured below the horizontal)
Summary
* Time of flight: 7.49 seconds
* Horizontal range: 337.05 meters
* Final velocity: 86.5 m/s at an angle of approximately 58.1° below the horizontal.
