Question

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

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.

Learn more about maximum height here: brainly.com/question/13665920

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;

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

Learn more about vertical velocity here: brainly.com/question/24949996