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.
(a) How much time passed while the ball was in the air?
s
(b) How far did the ball travel horizontally before landing?
m
(c) What was the peak height reached by the ball?
m
vyi = 37.0 sin52,8° = 29.47 m/s
vix = 37.0 cos52.8° = 22.37 m/s
(a) the ball's height y is
y(t) = 29.47t - 4.9t^2
So, when it hits the green,
29.47t - 4.9t^2 = 5.3
t = 5.829
(b) the horizontal distance, at constant speed is
x = 5.829 * 22.37 = 130.39
(c) the vertex of the parabola is at
t = -b/2a = 29.47/9.8 = 3.01
y(3.007) = 44.31
To solve this problem, we can use the equations of motion for projectile motion. Let's break down the problem step-by-step:
Step 1: Calculate the time of flight.
The time of flight (t) is the duration the ball spends in the air.
We can use the equation:
yf = yi + v0y * t + (1/2) * g * t^2, where yi = 0 (initial position) and vf = 0 (final velocity as the ball stops immediately).
Simplifying the equation, we have:
yf = v0y * t + (1/2) * g * t^2
Since the ball lands at the same horizontal position as it started, we know that v0x (horizontal component of initial velocity) does not change. Therefore, v0x * t = xf, where xf = horizontal distance traveled.
Step 2: Calculate the horizontal distance traveled (xf).
xf = v0x * t
Step 3: Calculate the vertical distance traveled (yf).
yf = yi + v0y * t + (1/2) * g * t^2
Step 4: Calculate the time of flight (t).
Substitute yi = 0, vf = 0, and solve for t:
yf = (v0 * sinθ) * t - (1/2) * g * t^2
Simplifying the equation, we have:
0.5 * g * t^2 - (v0 * sinθ) * t + yf = 0
Using the quadratic formula, we can solve for t:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
where a = 0.5 * g, b = -(v0 * sinθ), and c = yf.
Step 5: Calculate the peak height (ymax).
The peak height (ymax) is the highest point reached by the ball. It occurs halfway through the time of flight. Therefore, ymax = v0y * t / 2.
Now, let's calculate the answers for each part of the problem:
(a) How much time passed while the ball was in the air?
From step 4, we calculated that t = (-b ± sqrt(b^2 - 4ac)) / (2a).
Substituting the values:
a = 0.5 * g, b = -(v0 * sinθ), and c = yf,
we can calculate the time (t) from the positive root as the negative root would be unrealistic.
(b) How far did the ball travel horizontally before landing?
We can calculate the horizontal distance traveled using the equation xf = v0x * t, where v0x = v0 * cosθ.
(c) What was the peak height reached by the ball?
We can calculate the peak height using the equation ymax = v0y * t / 2, where v0y = v0 * sinθ.
Let's calculate each part using the given values:
θ = 52.8°,
v0 = 37.0 m/s,
yf = 5.3 m,
g = 9.8 m/s^2 (acceleration due to gravity).
(a) Time of flight:
a = 0.5 * g = 0.5 * 9.8 = 4.9 m/s^2
b = -(v0 * sinθ) = -(37.0 * sin(52.8°)) = -28.5905 m/s
c = yf = 5.3 m
Substituting these values into the quadratic formula, we get:
t = (-(-28.5905) ± sqrt((-28.5905)^2 - 4 * 0.5 * 9.8 * 5.3)) / (2 * 0.5 * 9.8)
t = (28.5905 ± sqrt(820.3465 + 105.28)) / 9.8
t = (28.5905 ± sqrt(925.6265)) / 9.8
Taking the positive root, we have:
t = (28.5905 + sqrt(925.6265)) / 9.8
Solving this equation, we get the time of flight.
(b) Horizontal distance traveled:
xf = v0x * t
Substituting the values:
v0x = v0 * cosθ,
v0 = 37.0 m/s,
θ = 52.8°,
and t calculated in part (a),
we can calculate the horizontal distance traveled.
(c) Peak height:
ymax = v0y * t / 2
Substituting the values:
v0y = v0 * sinθ,
v0 = 37.0 m/s,
and θ = 52.8°,
we can calculate the peak height reached.
Please provide the values, and I will calculate the answers for you.
To answer these questions, we can use the equations of motion for projectile motion. These equations are derived from the kinematic equations, but applied specifically to objects that are launched into the air and follow a curved path.
Let's break down each question and explain how to solve it:
(a) How much time passed while the ball was in the air?
To find the time the ball was in the air, we can use the equation for the vertical motion of the projectile:
y = yo + vot + (1/2)gt^2
In this equation:
- y is the final vertical position of the ball (yf = 5.3 m)
- yo is the initial vertical position of the ball (yo = 0 m)
- vo is the initial vertical velocity of the ball, which can be calculated using the initial speed and the launch angle (vo = v0 * sin(θ))
- g is the acceleration due to gravity (approximated as 9.8 m/s^2)
- t is the time taken for the ball to reach the final vertical position
Plugging in the values we know, the equation becomes:
5.3 = 0 + (37.0 * sin(52.8)) * t - (1/2)(9.8)(t^2)
Solving this equation will give us the time the ball was in the air.
(b) How far did the ball travel horizontally before landing?
To find the horizontal distance traveled by the ball, we can use the equation for the horizontal motion of the projectile:
x = xo + vot
In this equation:
- x is the horizontal distance traveled by the ball before landing
- xo is the initial horizontal position of the ball (xo = 0 m)
- vo is the initial horizontal velocity of the ball, which can be calculated using the initial speed and the launch angle (vo = v0 * cos(θ))
Substituting the values we know, the equation becomes:
x = 0 + (37.0 * cos(52.8)) * t
Solving this equation will give us the horizontal distance the ball traveled.
(c) What was the peak height reached by the ball?
The peak height reached by the ball occurs when its vertical velocity becomes zero. So, to find the peak height, we need to find the time at which this occurs. We can do this by finding the t value when the vertical velocity becomes zero.
The equation for vertical velocity can be given as:
v = vo + gt
In this equation:
- v is the vertical velocity of the ball
- vo is the initial vertical velocity of the ball, which can be calculated using the initial speed and the launch angle (vo = v0 * sin(θ))
- g is the acceleration due to gravity (approximated as 9.8 m/s^2)
- t is the time taken for the ball to reach the peak height
Setting v = 0 and solving for t will give us the time at which the ball reaches its peak height. We can then substitute this t value into the equation for vertical displacement to find the peak height.
By following these steps and plugging in the given values, we can find the answers to each question.