Question

6\%) Problem 5: A stationary soccer ball of mass m=0.54 kg is kicked with a constant force of F=26 N. The player's foot is in contact with the ball for t= .24 s. A 33\% Part (a) Write an expression for the speed of the ball, vi​, as it leaves the player's foot. a 33\% Part (b) What is the velocity of the ball right after contact with the foot of the player? a 33% Part (c) If the ball left the player's foot at an angle θ=45∘ relative to the horizontal, how high h did it go in meters?

Answer 1

The speed of the ball as it leaves the player's foot is 11.56 m/s. The velocity of the ball right after contact with the foot of the player is 0 m/s. The ball goes to a height of 3.52 meters if it left the player's foot at an angle of 45 degrees relative to the horizontal.

Step-by-step explanation:

Part (a):

For part (a), we need to find the expression for the speed of the ball, vi, as it leaves the player's foot. We can use the equation:

F = m·a

Solving for acceleration (a):

a = F/m

Given that F = 26 N and m = 0.54 kg, we have:
a = 26 N / 0.54 kg = 48.15 m/s2

Since the initial velocity is 0 m/s, we can use the equation:

vi = a·t

Substituting the values, we get:

vi = 48.15 m/s2 · 0.24 s = 11.56 m/s

Therefore, the speed of the ball as it leaves the player's foot is 11.56 m/s.

Part (b):

For part (b), we need to find the velocity of the ball right after contact with the foot of the player. Since the initial velocity is 0 m/s, the velocity right after contact will also be 0 m/s. Therefore, the velocity of the ball right after contact with the foot of the player is 0 m/s.

Part (c):

For part (c), we need to find the height h to which the ball goes if it left the player's foot at an angle of 45 degrees relative to the horizontal. To solve this, we can use the equation:

h = (vi2 · sin2(θ)) / (2 · g)

Substituting the values, we get:

h = (11.56 m/s)2 · sin2(45 degrees) / (2 · 9.8 m/s2) = 3.52 m

Therefore, the ball goes to a height of 3.52 meters if it left the player's foot at an angle of 45 degrees relative to the horizontal.

Answer 2

The expression for the speed of the ball as it leaves the player's foot is approximately 11.56 m/s. The velocity of the ball right after contact is the same. Considering the vertical motion, the ball reaches a maximum height of approximately 3.40 meters. These calculations are based on the given values for force, mass, time of contact, launch angle, and the acceleration due to gravity.

Part (a): Expression for the speed of the ball:

The acceleration of the ball can be calculated using the formula:

a = F / m

where:

a is the acceleration of the ball (m/s²)

F is the force applied by the player's foot (N)

m is the mass of the ball (kg)

Plugging in the given values:

a = 26 N / 0.54 kg ≈ 48.15 m/s²

The final velocity of the ball can be found using the formula:

v_f = v_i + a * t

where:

v_f is the final velocity of the ball (m/s)

v_i is the initial velocity of the ball (in this case, 0 m/s since it's stationary)

a is the acceleration of the ball (previously calculated)

t is the time of contact between the foot and the ball (0.24 s)

Therefore, the expression for the final velocity (the speed of the ball as it leaves the foot) is:

v_f = 0 m/s + 48.15 m/s² * 0.24 s ≈ 11.56 m/s

Part (b): Velocity of the ball after contact:

The velocity of the ball right after contact with the player's foot is the same as the final velocity calculated in part (a), which is approximately 11.56 m/s.

Part (c): Height reached by the ball:

To find the maximum height reached by the ball, we need to consider its vertical motion. The vertical component of the initial velocity is:

v_y = v_f * sin(θ)

where:

v_y is the vertical component of the initial velocity (m/s)

v_f is the final velocity (previously calculated)

θ is the launch angle (45°)

Plugging in the values:

v_y = 11.56 m/s * sin(45°) ≈ 8.16 m/s

At the peak of its trajectory, the ball's vertical velocity will be 0. Using the equations of motion, we can find the time it takes to reach the peak:

t_peak = v_y / g

where:

t_peak is the time to reach the peak (s)

g is the acceleration due to gravity (9.81 m/s²)

Substituting:

t_peak = 8.16 m/s / 9.81 m/s² ≈ 0.83 s

Finally, the maximum height reached by the ball can be calculated using the formula:

h = v_y * t_peak + 0.5 * g * t_peak²

where:

h is the maximum height reached (m)

Plugging in:

h = 8.16 m/s * 0.83 s + 0.5 * 9.81 m/s² * 0.83 s² ≈ 3.40 m

Therefore, the ball reaches a maximum height of approximately 3.40 meters.