A projectile is fired with an initial speed of 1000 m/s at an angle of 37 degrees above the horizontal. If air resistance is neglected, the horizontal component of the projectile's velocity after 20 s is approximately
A) 600 m/s
B) 800 m/s
C) 640 m/s
D) 40 m/s
E) 160 m/s
please help to solve this problem! thanks
Vf = Vo + g*t.
Vf = 1000 * cos37 + 9.8 * 20,
Vf = 798.6 + (-196) = 602.6 m / s.
Well, well, well, looks like we have a projectile on our hands! Don't worry, I'll help you with the math, but in my own funny way!
First, let's break this down. We have an initial speed of 1000 m/s and an angle of 37 degrees. The horizontal component of the velocity is what we're after. Let's call it Vx.
Now, the horizontal component of velocity remains constant throughout the projectile's motion. So, after 20 seconds, Vx will still be the same as it was at the beginning.
But wait a minute! We haven't found Vx yet! Silly me, let's calculate it.
Vx = initial velocity * cos(angle)
Vx = 1000 m/s * cos(37°)
Now, it's time to whip out your calculator. Punch in those numbers and let's solve this equation!
Vx ≈ 1000 m/s * 0.7986
Drumroll, please...
Vx ≈ 798.6 m/s
Hey, look at that, we found our horizontal component of velocity! So, the answer is B) 800 m/s! Woohoo!
Now, go impress your friends with your projectile knowledge. Just don't go launching any watermelons towards my face!
To solve this problem, we need to analyze the motion of the projectile and calculate the horizontal component of its velocity after 20 seconds.
First, let's break down the initial velocity into its horizontal and vertical components.
Given:
Initial speed (v₀) = 1000 m/s
Launch angle (θ) = 37 degrees
The horizontal component of the initial velocity (v₀x) can be calculated using cosine function:
v₀x = v₀ * cos(θ)
Now, let's calculate the vertical component of the initial velocity (v₀y) using the sine function:
v₀y = v₀ * sin(θ)
Next, we need to analyze the motion of the projectile. Since air resistance is neglected, the horizontal component of the velocity remains constant throughout the motion. This means that the horizontal component of the velocity after 20 seconds (vfx) will be the same as the initial horizontal component (v₀x).
Therefore, the horizontal component of the projectile's velocity after 20 seconds is approximately 1000 * cos(37) = 800 m/s.
So, the correct answer is B) 800 m/s.
To solve this problem, we need to break down the initial velocity of the projectile into its horizontal and vertical components.
The horizontal component of the initial velocity can be found using the formula:
Vx = V * cos(theta)
where Vx represents the horizontal component of velocity, V represents the initial velocity (1000 m/s), and theta represents the angle of elevation (37 degrees).
Similarly, the vertical component of the initial velocity can be found using the formula:
Vy = V * sin(theta)
where Vy represents the vertical component of velocity.
Now, let's calculate the horizontal and vertical components of the initial velocity:
Vx = 1000 * cos(37)
Vx ≈ 800 m/s
Vy = 1000 * sin(37)
Vy ≈ 600 m/s
Since air resistance is neglected, the horizontal component of velocity remains constant throughout the projectile's motion. Thus, the horizontal component of velocity after 20 seconds is still approximately 800 m/s.
Therefore, the answer is Option B) 800 m/s.
CORRECTION:
Vf = 1000 + (-9.8) * 20 = 1000 -196 = 804 @ 37 deg.
Vf(hor) = 804cos37 = 642 m /s.