a football is kicked at a velocity of 15m/s at an angel of 25 to the horizontal. calculate a)the total fligh time b)the maximum height c)the range of ball
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A Football Is Kicked At Avelocity Of 15m/s at An Angle Of 25 Horizontal Calculate:
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This handy article has all the formulas you will need.
https://en.wikipedia.org/wiki/Trajectory
1.9m
2 meter
To solve this problem, we can use the equations of motion under constant acceleration.
a) Total Flight Time:
The total flight time of an object can be calculated by dividing the total vertical displacement by the vertical component of velocity.
Given:
Initial velocity, u = 15 m/s
Launch angle, θ = 25 degrees
We need to find the vertical component of velocity (Vy) using trigonometry:
Vy = u * sin(θ)
Vy = 15 * sin(25°)
Vy ≈ 6.38 m/s
The total vertical displacement (H) can be calculated using the kinematic equation for vertical motion:
H = (Vy^2) / (2 * g)
where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
H = (6.38^2) / (2 * 9.8)
H ≈ 2.06 m
Now, we can calculate the total flight time (T) using the equation:
T = 2 * (Vy / g)
T = 2 * (6.38 / 9.8)
T ≈ 1.30 seconds
Therefore, the total flight time of the football is approximately 1.30 seconds.
b) Maximum Height:
The maximum height (Hmax) can be found using the kinematic equation for vertical motion:
Hmax = (Vy^2) / (2 * g)
Hmax = (6.38^2) / (2 * 9.8)
Hmax ≈ 2.06 m
Therefore, the maximum height of the football is approximately 2.06 meters.
c) Range:
The range (R) of the ball can be calculated using the horizontal component of the initial velocity (Vx) and the total flight time (T).
Vx = u * cos(θ)
Vx = 15 * cos(25°)
Vx ≈ 13.57 m/s
R = Vx * T
R = 13.57 * 1.30
R ≈ 17.64 meters
Therefore, the range of the ball is approximately 17.64 meters.