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.
Sure! To find the horizontal distance traveled by the football after 0.5 seconds, we can use the kinematic equations of motion. Let's break down the problem step by step:
Step 1: Resolve the initial velocity into horizontal and vertical components.
The initial velocity can be split into two components: the horizontal component (Vx) and the vertical component (Vy). We can find these components using trigonometry.
Given: Initial velocity (Vi) = 42 ft/sec, Angle (θ) = 35 degrees
Vx = Vi * cos(θ)
= 42 ft/sec * cos(35 degrees)
= 42 ft/sec * 0.819
≈ 34.4 ft/sec
Vy = Vi * sin(θ)
= 42 ft/sec * sin(35 degrees)
= 42 ft/sec * 0.573
≈ 24 ft/sec
Step 2: Calculate the horizontal distance traveled (x) after 0.5 seconds.
We know that the horizontal motion of the football is independent of the vertical motion. Therefore, we can calculate the distance traveled horizontally using the formula:
x = Vx * t
= 34.4 ft/sec * 0.5 sec
= 17.2 ft
Therefore, the football has traveled approximately 17.2 feet horizontally after 0.5 seconds.
Remember to round your answer to the nearest tenth, so the final answer would be approximately 17.2 feet.