A football is kicked with an initial velocity of 42 ft/sec at an angle of 35 degrees with the horizontal. How far has the football traveled horizontally after 0.5 sec? round to nearest tenth.

Could someone explain this please?

x=t|v|cos(x)
x=t(42)cos35
x=42tcos35

y=t|v|sin(x)-1/2gt^2+h
y=t(42)sin35-1/2(32)t^2+0.5
y=42tsin35-16t^2+0.5

0.5=42tcos35
t=0.5/42cos35
t=0.00975181

I don't know, I just followed an example in my book

Let me edit your work:

HORIZONTAL>>
x=t|v|cos(x)
x=t(42)cos35
x=42tcos35

so this is the equation of the horizontal component of the football. At time 1/2 second, put that in for t, and you have the horizontal distance.

The rest of your work is not related.

17.2ft?

To solve this problem, you can use the equations of motion for projectile motion. Let's break it down step by step:

1. Recall that the horizontal and vertical components of the initial velocity can be found using trigonometry:

Horizontal component: Vx = |V| * cos(θ)
Vertical component: Vy = |V| * sin(θ)

Given:
Initial velocity, V = 42 ft/sec
Angle with the horizontal, θ = 35 degrees

Hence:
Vx = 42 * cos(35)
Vx ≈ 34.252 ft/sec

2. Next, we can find the time it takes for the football to travel horizontally for 0.5 seconds. Considering only the horizontal motion, we can write:

x = Vx * t
where x is the horizontal distance traveled after time t.
Given:
t = 0.5 seconds

Hence:
x = 34.252 * 0.5
x ≈ 17.126 ft

Therefore, the football has traveled approximately 17.1 feet horizontally after 0.5 seconds.

Note: The steps above assume no air resistance and a flat surface. The value of acceleration due to gravity used is rounded to 32 ft/sec^2 for simplicity.