A golf ball is it with the initial velocity of 50.0 m/s at an angle of 55 degrees above the horizontal. (a) How high will the ball go? (b) What is the total time the ball is in the air? (c) How far will the ball travel horizontally before it hits the ground?
STEP 1:
Determine horizontal(x) and vertical (y) components of the initial velocity of golf ball.
Vy=Vsin55 = 50x.8191= 40.9 m/s, Vx=Vcos55 = 50x.5735= 28.6 m/s
STEP 2:
At maximum height, Vy=0.
Determine time (t) it takes to reach max. height.
Vy(final)=Vy(initial)+(-g)x(t)
Solve for (t):
0=40.9 x (-9.8) x t
0=40.9 x (-9.8 t)
9.8t=40.9, t=40.9/9.8, t= 4.17 sec
Since this half the total time in air, multiply by 2,
= 8.3 SEC (TOTAL T IN AIR)
STEP 3:
DETERMINE HIGHEST POINT REACHED BY BALL
h=((V**2)(Sin**2 55))/ 2g
h=((50**2) (.8191**2)) / 2 (9.8)
h=((2500)(.6710))/19.6
h= 85m Height reached
STEP 4:
DETERMINE 'RANGE' (total horizontal distance traveled)
R=Vx (t)
R=28.6 x 8.3
R=237.38 meters
thank you very much for your help M Bil
To solve this problem, we can use the equations of motion for projectile motion. Let's break it down step-by-step:
Step 1: Analyze the given information
Initial velocity (v₀) = 50.0 m/s
Launch angle (θ) = 55 degrees
Step 2: Find the vertical component of the initial velocity (v₀y)
v₀y = v₀ * sin(θ)
v₀y = 50.0 m/s * sin(55°)
v₀y ≈ 40.58 m/s
Step 3: Find the time taken to reach maximum height (tₘ)
The time to reach maximum height is equal to the time taken for the vertical component of velocity to become zero.
tₘ = v₀y / g
(Note: g is the acceleration due to gravity, approximately 9.8 m/s²)
tₘ = 40.58 m/s / 9.8 m/s²
tₘ ≈ 4.15 s
Step 4: Find the maximum height (h)
h = (v₀y²) / (2 * g)
h = (40.58 m/s)² / (2 * 9.8 m/s²)
h ≈ 91.42 m
Therefore, the ball will reach a maximum height of approximately 91.42 meters.
Step 5: Find the total time of flight (T)
Since the total time of flight is the time taken to reach the maximum height (tₘ) plus the time taken to return to the ground (2 * tₘ), we can calculate T as follows:
T = 2 * tₘ
T = 2 * 4.15 s
T ≈ 8.30 s
Therefore, the total time the ball is in the air is approximately 8.30 seconds.
Step 6: Find the horizontal distance traveled (R)
The horizontal distance traveled by the ball can be calculated using the horizontal component of the initial velocity (v₀x) and the total time of flight (T).
v₀x = v₀ * cos(θ)
v₀x = 50.0 m/s * cos(55°)
v₀x ≈ 26.51 m/s
R = v₀x * T
R = 26.51 m/s * 8.30 s
R ≈ 220.48 m
Therefore, the ball will travel approximately 220.48 meters horizontally before it hits the ground.
Summary:
(a) The ball will reach a maximum height of approximately 91.42 meters.
(b) The total time the ball is in the air is approximately 8.30 seconds.
(c) The ball will travel approximately 220.48 meters horizontally before it hits the ground.
To find the answers to these questions, we can use the equations of projectile motion. Projectile motion refers to the motion of objects that are launched into the air and then move under the influence of gravity. In this case, the golf ball is launched with an initial velocity, given an angle above the horizontal.
Let's break down each part of the question and solve it step by step:
(a) How high will the ball go?
To find the maximum height the ball will reach, we need to find the vertical component of the initial velocity. We can use the equation:
Vertical component of velocity (Vy) = initial velocity (V) * sin(angle)
Vy = 50.0 m/s * sin(55 degrees)
Now, we can use this value to find the time it takes for the ball to reach its maximum height. We use the equation:
Time of flight (t) = 2 * (Vertical component of velocity) / acceleration due to gravity (g)
t = 2 * Vy / g
Note: the acceleration due to gravity is usually considered as -9.8 m/s^2 (negative because it acts downwards).
Next, we can use the time of flight to find the maximum height (H) using the equation:
Maximum height (H) = (Vertical component of velocity)^2 / (2 * g)
H = Vy^2 / (2 * g)
(b) What is the total time the ball is in the air?
The total time in the air is twice the time it takes to reach the maximum height since the time of ascent is equal to the time of descent. Therefore,
Total time in the air = 2 * time of flight (t)
(c) How far will the ball travel horizontally before it hits the ground?
To determine the horizontal distance traveled (R), we need to find the horizontal component of the initial velocity. We can use the equation:
Horizontal component of velocity (Vx) = initial velocity (V) * cos(angle)
Vx = 50.0 m/s * cos(55 degrees)
Now, we can use this value and the total time in the air to find the horizontal distance traveled (R) using the equation:
Horizontal distance (R) = horizontal component of velocity (Vx) * total time in the air
R = Vx * (2 * t)
So, to recap:
(a) To find the maximum height (H):
- Calculate the vertical component of velocity (Vy = V * sin(angle))
- Use the equation H = Vy^2 / (2 * g) to find the maximum height.
(b) To find the total time in the air (t):
- Calculate the time of flight (t) using the equation t = 2 * Vy / g
- The total time in the air is 2 * t.
(c) To find the horizontal distance traveled (R):
- Calculate the horizontal component of velocity (Vx = V * cos(angle))
- Use the equation R = Vx * (2 * t) to find the horizontal distance traveled.