Question

A ball is kicked a horizontal distance of 45.8 m. If it reaches a maximum height of 24.2 m with a flight time of 4.4 seconds, was the ball kicked at a projection angle less than, greater than, or equal to 45°? Provide a rationale for your answer based on the appropriate calculations.

Answer 1

By analyzing the projectile motion of the ball and using kinematics equations to find the initial vertical and horizontal velocity components, we can calculate a launch angle of approximately 63.44°. Therefore, the ball was kicked at an angle greater than 45°.

Step-by-step explanation:

To determine whether the ball was kicked at an angle less than, greater than, or equal to 45°, we analyze the projectile motion of the ball. Given the flight time of 4.4 seconds and a maximum height of 24.2 m, we can use the kinematic equation for vertical motion (h = v_y * t - (1/2) * g * t^2) to find the initial vertical velocity component (v_y). Since the ball reaches its maximum height at half of the total flight time, we can plug in 2.2 seconds (half of 4.4 seconds) for the time, and 24.2 m for the maximum height reached to solve for the initial vertical component of velocity:

v_y = (h + (½) * g * t^2) / t = (24.2 + (0.5) * 9.8 * 2.2^2) / 2.2 = 20.6 m/s, where g is the acceleration due to gravity (9.8 m/s^2).

Now, we apply the horizontal motion kinematics (d = v_x * t) to find the initial horizontal velocity component (v_x). We have the total horizontal distance (45.8 m) and the total time of flight (4.4 s):

v_x = d / t = 45.8 / 4.4 = 10.41 m/s.

Now, we can determine the launch angle (θ) using the trigonometric relation tan(θ) = v_y / v_x. Calculating the angle gives:

θ = tan⁻¹ (v_y / v_x) = tan⁻¹ (20.6 / 10.41) = 63.44°, which means the ball was kicked at an angle greater than 45°.

Answer 2

Answer: A soccer ball is kicked with an initial horizontal speed of 5 m/s and an initial vertical speed of

3 m/s. Assuming that projection and landing height are the same and neglecting air resistance,

identify the following quantities:

a. The ball’s horizontal speed is 0.5 seconds into its flight

5 m/s because horizontal speed is constant

b. The ball’s horizontal speed midway through its flight

5 m/s because horizontal speed is constant

c. The ball’s horizontal speed immediately before contact with the ground

5 m/s because horizontal speed is constant

d. The ball’s vertical speed at the apex of its flight

0 m/s because the ball changes directions at the apex and must reach zero speed

before changing directions

e. The ball’s vertical speed midway through its flight

0 m/s – the apex is midway because the projection and landing height are the same, so

the reason is the same as d.

f. The ball’s vertical speed immediately before contact with the ground

3 m/s because the projection and landing heights are the same, projection and landing

speeds are the same

Step-by-step explanation: