a football is kicked with an initial speed of 80 m/s at an angle of 30 degrees above the horizontal.
A. what is the maximum height that the ball attains?
b. what is the balls hang time? (how long is it in the air)
c. what is the range of the ball? (how far did it go)
Vo = 80m/s[30o].
Xo = 80*Cos30 = 69.3 m/s.
Yo = 80*sin30 = 40 m/s.
a. Y = Yo + g*Tr.
0 = 40 - 9.8Tr, Tr = 4.08 s. = Rise time.
h = Yo*Tr + 0.5g*Tr^2.
h = 40*4.08 -4.9*4.08^2 = 81.6 m.
b. Tf = Tr = 4.08 m/s. = Fall time.
Tr+Tf = 4.08 + 4.08 = 8.16 s. = Time in air.
c. Range = Xo*(Tr+Tf) = 69.3 * 8.16 = 565.5 m.
To solve these problems, we can use the basic kinematic equations of projectile motion. Let's start by breaking down the initial velocity into its horizontal and vertical components.
Given:
Initial speed (Vi) = 80 m/s
Launch angle (θ) = 30 degrees
Step 1: Calculate the initial vertical velocity (Vy):
Vy = Vi * sin(θ)
Vy = 80 * sin(30)
Vy ≈ 40 m/s
Step 2: Calculate the initial horizontal velocity (Vx):
Vx = Vi * cos(θ)
Vx = 80 * cos(30)
Vx ≈ 69.28 m/s
Now, let's solve each part of the problem:
A. Maximum height:
To find the maximum height, we need to determine the time it takes for the ball to reach the highest point. At this point, the vertical velocity component becomes zero (Vf = 0).
The equation we'll use is:
Vf = Vy - gt
0 = Vy - gt
Solving for time (t):
t = Vy / g
t ≈ 40 / 9.8
t ≈ 4.08 seconds
Now that we have the time, we can find the maximum height (h):
h = Vy * t - 0.5 * g * t^2
h = 40 * 4.08 - 0.5 * 9.8 * (4.08)^2
h ≈ 163.2 - 82.03
h ≈ 81.17 meters
Therefore, the maximum height that the ball attains is approximately 81.17 meters.
B. Hang time:
The hang time refers to the amount of time it takes for the ball to go up and come back down, which is double the time it took to reach the highest point.
Hang time (T) = 2 * t
T ≈ 2 * 4.08
T ≈ 8.16 seconds
Therefore, the ball is in the air for approximately 8.16 seconds.
C. Range:
The range refers to the distance the ball travels horizontally. To find it, we need to calculate the horizontal displacement (D).
D = Vx * T
D ≈ 69.28 * 8.16
D ≈ 566.54 meters
Therefore, the range of the ball is approximately 566.54 meters.