A golf ball was hit and projected at an angle of 75 degree with the horizontal. If the initial velocity of the ball was 40 m/s
A. time t the golf ball was on air, find the following
B. horizontal distance dx the ball travelled
C. maximum height dy the ball reached.
To find the answers to the given questions, we can use the equations of motion for projectile motion.
A. To find the time the golf ball was in the air (t), we can use the equation:
t = 2 * (v * sin(θ)) / g
Where,
v = initial velocity of the ball = 40 m/s
θ = angle of projection = 75 degrees
g = acceleration due to gravity = 9.8 m/s^2
Using substitution:
t = 2 * (40 * sin(75)) / 9.8
t ≈ 5.42 seconds
So, the golf ball was in the air for approximately 5.42 seconds.
B. To find the horizontal distance the ball traveled (dx), we can use the equation:
dx = v * cos(θ) * t
Using substitution:
dx = 40 * cos(75) * 5.42
dx ≈ 132.8 meters
So, the ball traveled approximately 132.8 meters horizontally.
C. To find the maximum height the ball reached (dy), we can use the equation:
dy = (v^2 * sin^2(θ)) / (2 * g)
Using substitution:
dy = (40^2 * sin^2(75)) / (2 * 9.8)
dy ≈ 52.1 meters
So, the ball reached a maximum height of approximately 52.1 meters.