A golf ball is hit with an initial velocity of 50 m/s at an angle of 45° above the horizontal. What's the maximum height it will reach?
The vertical component of the velocity is v=50sin(45°).
Ignore air resistance and equate energy:
(1/2)mv² = mgh
h=v²/(2g)
Well, let me do some quick calculations while trying to not get teed off here. Now, when a golf ball is launched, it follows a lovely parabolic trajectory. At the highest point of this trajectory, the vertical component of the ball's velocity becomes zero. Ah, the magic of physics! So, let's get to the point and determine the maximum height.
To find the maximum height, we need to break down the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion because there are no external horizontal forces acting on the ball. Since the angle is 45°, the vertical and horizontal components will be equal, each being 50 m/s multiplied by the cosine of 45°.
The vertical velocity, on the other hand, changes due to gravity. To reach the maximum height, the vertical velocity must become zero. So, we can say that:
Vertical component of velocity = 0
Now, we can find the time it takes for the ball to reach its maximum height. Knowing that the initial vertical velocity is equal to the final vertical velocity (which is zero), and using the equation v = u + at (where v is final velocity, u is initial velocity, a is acceleration, and t is time), we can calculate the time of flight.
Starting with an initial vertical velocity of 50 m/s multiplied by the sine of 45°, and with an acceleration due to gravity of 9.8 m/s^2 (I hope you're not feeling too down), we can solve for time.
v = u + at
0 = (50 × sin 45°) + (-9.8 × t)
Simplifying this equation, we find that t = (50 × sin 45°) / 9.8
Having the time it takes to reach maximum height, we can now calculate the vertical displacement using the equation s = ut + (1/2)at^2 (where s is displacement).
s = (50 × sin 45°) × [(50 × sin 45°) / 9.8] + (1/2) × (-9.8) × [(50 × sin 45°) / 9.8]^2
Crunching the numbers (or rather, letting a calculator crunch them for us), we find that the maximum height the golf ball will reach is approximately 127.55 meters above the ground. That's quite an elevated swing! Keep in mind that these calculations are done in the ideal world of physics, assuming no air resistance and all that jazz.
To find the maximum height the golf ball will reach, we need to analyze the projectile motion of the ball.
Step 1: Resolve the initial velocity into horizontal and vertical components.
The initial velocity of the golf ball is given as 50 m/s, and it is at an angle of 45° above the horizontal. We can resolve this velocity into its horizontal (Vx) and vertical (Vy) components using trigonometry.
Vx = V * cos(θ)
Vx = 50 m/s * cos(45°) ≈ 35.35 m/s
Vy = V * sin(θ)
Vy = 50 m/s * sin(45°) ≈ 35.35 m/s
Step 2: Determine the time it takes for the ball to reach its maximum height.
At the maximum height, the vertical component of the velocity becomes zero (Vy = 0 m/s). We can use this information to find the time it takes for the ball to reach its maximum height using one of the kinematic equations:
Vy = V0y + ay * t
0 = 35.35 m/s + (-9.8 m/s^2) * t
Solving for t:
35.35 m/s = 9.8 m/s^2 * t
t ≈ 3.61 s
Step 3: Calculate the maximum height.
To find the maximum height, we can use the following kinematic equation:
Δy = V0y * t + (1/2) * ay * t^2
Δy = 35.35 m/s * 3.61 s + (1/2) * (-9.8 m/s^2) * (3.61 s)^2
Δy ≈ 63.7 m
Therefore, the maximum height the golf ball will reach is approximately 63.7 meters.
To find the maximum height a golf ball will reach, we can use the equations of motion for projectile motion.
Step 1: Break the initial velocity into its horizontal and vertical components.
The initial velocity (v₀) of 50 m/s can be broken down into its horizontal and vertical components using trigonometry.
The horizontal component (v₀x) remains constant throughout the motion and can be calculated using v₀x = v₀ * cosθ, where θ is the angle of 45°.
The vertical component (v₀y) can be calculated using v₀y = v₀ * sinθ.
Given:
v₀ = 50 m/s
θ = 45°
v₀x = v₀ * cosθ
= 50 * cos(45°)
= 50 * 0.7071
≈ 35.355 m/s
v₀y = v₀ * sinθ
= 50 * sin(45°)
= 50 * 0.7071
≈ 35.355 m/s
Step 2: Determine the time it takes for the ball to reach its maximum height.
At the highest point, the vertical component of velocity becomes zero (v_y = 0). We can use the equation v_y = v₀y - g * t, where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time.
0 = v₀y - g * t
Solving for t, we have t = v₀y / g.
t = (35.355 m/s) / (9.8 m/s²)
≈ 3.6 s
Step 3: Calculate the maximum height (h).
The maximum height can be determined using the equation h = v₀y * t - 0.5 * g * t².
h = (35.355 m/s) * (3.6 s) - 0.5 * (9.8 m/s²) * (3.6 s)²
≈ 63.14 meters
Therefore, the maximum height the golf ball will reach is approximately 63.14 meters.