How high will an object rise if it is projected upward at a 45 degree angle and an initial velocity of 50 m/s?
To determine how high an object will rise when projected upward at a 45-degree angle with an initial velocity of 50 m/s, we can follow these steps:
1. Break down the initial velocity into horizontal and vertical components. Since the angle of projection is 45 degrees, the initial velocity can be divided equally into its horizontal and vertical components. Using basic trigonometry, the vertical component equals the horizontal component:
Vx (horizontal) = 50 m/s * cos(45°)
Vx (horizontal) = 50 m/s * √2 / 2
Vx (horizontal) = 35.35 m/s
Vy (vertical) = 50 m/s * sin(45°)
Vy (vertical) = 50 m/s * √2 / 2
Vy (vertical) = 35.35 m/s
2. Determine the time taken to reach the maximum height. When an object is projected vertically upward, its vertical velocity decreases until it becomes zero at the highest point. At this point, the object will start to descend. We can use the basic kinematic equation for vertical motion to find the time taken to reach the maximum height:
Vy (vertical) = Vy (initial) - g * t
0 m/s = 35.35 m/s - 9.8 m/s² * t
Solving for t:
35.35 m/s - 9.8 m/s² * t = 0
t = 3.61 seconds
3. Calculate the maximum height reached by the object. Once we have the time taken to reach the maximum height, we can use another kinematic equation to calculate it:
y = y (initial) + (Vy (initial) * t) - (0.5 * g * t²)
y = 0 + (35.35 m/s * 3.61 s) - (0.5 * 9.8 m/s² * (3.61 s)²)
Solving for y:
y ≈ 63.3 meters
Hence, the object will rise to approximately 63.3 meters when projected upward at a 45-degree angle with an initial velocity of 50 m/s.