A golfer is enjoying a day out on the links. What maximum height will a 345m drive reach if it is launched at an angle of 21 degrees to the ground? the acceleration due to gravity is 9.81m/s^2
Given the distance and angle of elevation,
the velocity with which the ball left the club face derives from
d = V^2(sin(µ))/g
which then allows the maximum height reached to be found from
h = [V^2(sin^2(µ))]/2g
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To find the maximum height reached by the golf ball, we can use the physics equation for projectile motion. The equation is:
h = (v₀² * sin²(θ)) / (2 * g),
where:
h is the maximum height reached,
v₀ is the initial velocity of the golf ball,
θ is the launch angle, and
g is the acceleration due to gravity.
In this case, we are given:
v₀ = 345 m,
θ = 21°, and
g = 9.81 m/s².
First, we need to calculate the initial vertical velocity (v₀ₓ) and the initial horizontal velocity (v₀ᵧ).
v₀ₓ = v₀ * sin(θ),
v₀ᵧ = v₀ * cos(θ).
Next, we can calculate the time it takes for the golf ball to reach its maximum height. We'll use the equation:
t = v₀ₓ / g.
Once we have the time (t), we can calculate the maximum height (h) using the formula mentioned earlier:
h = (v₀ₓ²) / (2 * g).
Now, let's plug in the values to find the answer:
v₀ₓ = 345 m * sin(21°),
v₀ₓ ≈ 119.687 m/s.
v₀ᵧ = 345 m * cos(21°),
v₀ᵧ ≈ 321.94 m/s.
t = v₀ₓ / g,
t = 119.687 m/s / 9.81 m/s²,
t ≈ 12.202 s.
h = (v₀ₓ²) / (2 * g),
h = (119.687 m/s)² / (2 * 9.81 m/s²),
h ≈ 719.83 m.
Therefore, the maximum height reached by the golf ball is approximately 719.83 meters.