Question

A soccer ball is kicked at an angle with an initial horizontal velocity 0,=18 m/s and an initial vertical velocity 0,=15 m/s, as shown in the figure. A soccer ball on the ground. Coordinate axes indicate that the X axis points right and the Y axis points up. An arrow points up and right from the ball at an angle theta above the horizontal. A dashed horizontal arrow labeled V subscript 0,x baseline points right from the tail of the slanted arrow to a point vertically below the tip of the slanted arrow. A dashed vertical arrow labeled V subscript 0,y baseline points up from the tip of V sub 0,x to the tip of the slanted arrow. What is the height ℎ(t) of the ball t=2.4 s after the kick?

Answer 1

198.47 meters

Step-by-step explanation:

To determine the height of the soccer ball at a given time, we need to break down the initial velocities into their horizontal and vertical components and then use the equations of motion.

Given: Initial horizontal velocity (V₀,x) = 18 m/s Initial vertical velocity (V₀,y) = 15 m/s Time (t) = 2.4 s

Step 1: Find the vertical displacement (Δy) using the equation: Δy = V₀,y * t + (1/2) * g * t²

Here, g represents the acceleration due to gravity, which is approximately 9.8 m/s².

Substituting the given values: Δy = (15 m/s) * (2.4 s) + (1/2) * (9.8 m/s²) * (2.4 s)²

Step 2: Calculate the height (h) by adding the initial height to the vertical displacement: h = 0 + Δy

Simplifying the equation: h = (15 m/s) * (2.4 s) + (1/2) * (9.8 m/s²) * (2.4 s)²

Now, let's calculate the height:

h = (36 m) + (1/2) * (9.8 m/s²) * (5.76 s²)

h = 36 m + (4.9 m/s²) * (5.76 s²)

h = 36 m + 4.9 m/s² * 33.1776 s²

h ≈ 36 m + 162.46944 m

h ≈ 198.47 m

Therefore, the height of the soccer ball at t = 2.4 s after the kick is approximately 198.47 meters