A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. How far did the ball travel before hitting the ground? Round the answer to the nearest tenth of a meter.
just to point out and make it aware, your all wrong. I found out the hard way when I failed my test. So y'all need to get with the program.
just guess
plug and chug: the Range R is
R = v^2/g sin2θ
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To find the distance the golf ball traveled before hitting the ground, we can use the equations of projectile motion.
First, we need to break the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity.
The horizontal component of the initial velocity can be found using the cosine function:
Vx = V₀ * cos(θ)
where:
Vx is the horizontal component of the velocity
V₀ is the initial velocity of the ball
θ is the launch angle
Substituting the values given in the question:
Vx = 31 m/s * cos(35°)
Next, we can calculate the time it takes for the ball to hit the ground. We'll use the following formula for the time of flight in projectile motion:
t = 2 * Vy / g
where:
t is the time of flight
Vy is the vertical component of the velocity
g is the acceleration due to gravity (approximately 9.8 m/s²)
The vertical component of the initial velocity can be found using the sine function:
Vy = V₀ * sin(θ)
Substituting the values:
Vy = 31 m/s * sin(35°)
Now we can calculate the time of flight:
t = 2 * (31 m/s * sin(35°)) / 9.8 m/s²
Finally, we can calculate the horizontal distance traveled by the ball using the equation:
distance = Vx * t
Let's plug in the values and calculate the answer:
Vx = 31 m/s * cos(35°)
Vy = 31 m/s * sin(35°)
t = 2 * (31 m/s * sin(35°)) / 9.8 m/s²
distance = Vx * t