A golfer hits his approach shot at an angle of
θ = 52.8°,
giving the ball an initial speed of
v0 = 37.0 m/s
(see figure below). The ball lands on the elevated green,
yf = 5.3 m
above the initial position near the hole, and stops immediately.
please delete this question. this the wrong incomplete one.
To find the range, which is the horizontal distance the ball travels, we need to find the time of flight. We can use the equation:
tf = 2 * tfmax
where tfmax is the time taken for the ball to reach its highest point. The formula for tfmax is:
tfmax = vy / g
where vy is the vertical component of the initial velocity and g is the acceleration due to gravity.
Given that the initial velocity, v0, makes an angle θ with the horizontal, we can find vy using:
vy = v0 * sin(θ)
Next, we can calculate tfmax using the above formula.
Now that we know tf, we can find the horizontal distance, or the range, using:
R = vx * tf
where vx is the horizontal component of the initial velocity.
To find vx, we use:
vx = v0 * cos(θ)
Now, we can find the range by substituting the values we have into the equation.
R = (v0 * cos(θ)) * (2 * (v0 * sin(θ)) / g)
Substitute the given values into the equation and calculate to find the range.