The given kinematic equations of motion are derived from Newton's Laws
of Motion.
Reasons:
The direction the ball is kicked = Upwards in the air
Maximum height to which the ball rises = 6.25 m
A) The speed of the ball the instant it left the soccer player's foot
Solution:
The equation to use is; v² = v₀² + 2·[tex]a_y[/tex]·(y - y₀)
[tex]a_y[/tex] = The acceleration due to gravity ≈ -9.81 m/s²
At the maximum height, y - y₀ = 6.25 m
The velocity at the maximum height, v = 0
v₀ = The speed of the ball the instant it left the soccer player's foot
Which gives;
v² = v₀² + 2·[tex]a_y[/tex]·(y - y₀)
0 = v₀² - 2 × 9.81 × (6.25) = v₀² - 122.625
v₀² = 122.625
v₀ = √(122.625) ≈ 11.074
The ball's speed the instant it left the soccer player's foot, v₀ ≈ 11.074 m/s.
B) Required:
The time it takes the soccer ball to rise to the maximum height.
Solution:
The equation of motion required is v = v₀ + [tex]a_y[/tex]·t
At the maximum height, v = 0
v₀ ≈ 11.074 m/s
v = v₀ + [tex]a_y[/tex]·t
Which gives;
0 = 11.074 - 9.81 × t
[tex]t \approx \dfrac{11.074}{9.81} \approx 1.129[/tex]
The time it took the ball to rise to its maximum height, t ≈ 1.129 seconds
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