A golf ball hit at a speed of 77 m/s will carry (achieve its maximum flight distance of) 256 m when driven at a launch angle of 16 deg. What would be its horizontal range in a vacuum if launched at the same speed and at the same angle ?
probably very close to the same value. A golf ball is designed to minimize drag.
However, if you have some data on air resistance to golf balls, I'm sure you can work it in somehow.
0 = (77sin16 - 4.9t)t
t=0 is trivial, find the other
x = 77cos16 * t
To find the horizontal range of a golf ball launched at a specific speed and angle in a vacuum, you can use the following formula:
Range = [(Initial Velocity)^2 * sin(2 * Angle)] / Gravity
In this case, the initial velocity is 77 m/s, the angle is 16 degrees, and we assume the gravity to be 9.8 m/s^2.
First, convert the angle from degrees to radians:
Angle in radians = (Angle in degrees) * (π / 180)
Angle in radians = 16 * (π / 180)
Next, plug the values into the formula to calculate the horizontal range:
Range = [(77)^2 * sin(2 * (16 * (π / 180)))] / 9.8
Using a scientific calculator, evaluate the equation:
Range ≈ 88.17 meters
Therefore, the horizontal range of the golf ball launched in a vacuum at the same speed and angle is approximately 88.17 meters.