Question

A kicked soccer ball has an initial velocity of 25 meters per second at an angle of 40 degrees, above the horizontal.

1) Calculate the magnitude of the vertical component of the ball's initial velocity.
2) Calculate the maximum height the ball reaches above its initial position.

Answer 1

(1) The vertical magnitude of the ball's initial velocity is 16.075 m/s,

(2) The maximum height of the ball is 13.18 m

(1) To Calculate the vertical component of the ball's initial velocity, we use the formula below.

Formula:

• Vy = Vsin∅........................ Equation 1

Where:

• Vy = vertical component of the ball's initial velocity
• V = Initial velocity of the ball
• ∅  = angle to the horizontal.

From the question,

Given:

• V = 25 m/s
• ∅ = 40°

Substitute these values into equation 1

• Vy = 25(sin40°)
• Vy = 25×0.643
• Vy = 16.075 m/s.

(2) To calculate the maximum height reached by the ball, we use the formula below.

Formula:

• H = (Vy)²/2g............... Equation 2

Where:

• g = Acceleration due to gravity = 9.8 m/s²

Substitute the value above into equation 2

• H = (16.075)²/(2×9.8)
• H = 13.18 m

Hence, (1) The vertical magnitude of the ball's initial velocity is 16.075 m/s, (b) The maximum height of the ball is 13.18 m.

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Answer 2

The vertical component of the ball's velocity is 16.1 m/s.

The maximum height the ball reaches above its initial position is 13.225 m.

The given parameters:

• Initial velocity of the ball, u = 25 m/s
• Angle of inclination, Ф = 40 degrees

What is vertical component of velocity?

The vertical component of velocity is the velocity in the y-direction.

The vertical component of the ball's velocity is calculated as follows;

$v_y = vsin(\theta)\\\\ v_y = 25 \times sin(40)\\\\ v_y = 16.1 \ m/s$

The maximum height the ball reaches above its initial position is calculated as follows;

$v_y_f^2 = v_y_i^2 -2gh\\\\ 0 = v_y_i^2 -2gh\\\\ 2gh = v_y_i^2 \\\\ h = \frac{v_y_i^2}{2g} \\\\ h = \frac{16.1^2 }{2 \times 9.8} \\\\ h = 13.225 \ m$

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