an arrow is shot at a 30 degree angle with the horizontal. it has a velocity of 49 m/s how high will the arrow go? what horizontal distance will the arrow travel? and how long will the arrow be in the air?
Vo = 49m/s[30o] = Initial velocity.
Xo = 49*Cos30 = 42.4 m/s = Ver. component.
Yo = 49*sin30 = 24.5 m/s = Hor. component.
A. Y^2 = Yo^2 + 2g*h.
0 = 24.5^2 + (-19.6)*h,
h = 30.63 meters.
B. Range = Vo^2*sin(2A)/g
Range = 49^2*sin(60)/9.8 = 212.2 m.
C. Y = Yo + g*Tr.
0 = 24.5 + (-9.8)Tr,
Tr = 2.5 s. = Rise time.
Tf = Tr = 2.5 s = Fall time.
Tr+Tf = 2.5 + 2.5 = 5.0 s. = Time in air.
Well, well, well, we have an arrow enthusiast here! Let me help you out with some clownish calculations.
First, let's break it down. Since the arrow is shot at a 30-degree angle, we know that it has both a vertical and horizontal component to its velocity.
Now, let's focus on the vertical component. To find out how high the arrow will go, we need to determine the time it takes for the arrow to reach its peak. The vertical velocity can be found using the formula v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
Since the arrow is shot at an angle, the vertical component of its initial velocity is given by u*sin(30) = 49*sin(30). This will help us find the time it takes for the arrow to reach its peak, which happens when the vertical velocity becomes zero.
To find the time, we can use the formula v = u + gt and set it equal to zero. Thus, 0 = 49*sin(30) + (-9.8)*t. Solving for t, we get t = (49*sin(30)) / 9.8.
Now, since the arrow reaches its peak halfway through its flight time, the total time in the air is double that. Thus, the total time in the air is 2 * (49*sin(30)) / 9.8.
To find the height the arrow reaches, we can use the formula h = u*t + 0.5*g*t^2. Plugging in the relevant values, we get h = (49*sin(30))*(49*sin(30))/(2*9.8).
As for the horizontal distance, we can use the formula d = u*cos(30)*t, where d is the horizontal distance, u is the initial velocity, t is the total time in the air, and theta is the launch angle.
Now, it's time for me to put on my clown glasses and calculate these values for you!
After the clownish calculations, the height the arrow will go is approximately (49*sin(30))^2 / (2*9.8) meters.
The horizontal distance the arrow will travel is approximately (49*cos(30)) * ((49*sin(30)) / 9.8) meters.
And finally, the overall time the arrow will be in the air is approximately 2 * ((49*sin(30)) / 9.8) seconds.
I hope my clownish calculations brought a smile to your face!
To solve this problem, we can break it down into two components: the horizontal motion and the vertical motion of the arrow.
Let's start by finding the time it takes for the arrow to reach its maximum height. We'll use the vertical component of the velocity for this calculation.
First, we need to find the initial vertical velocity component (Vy) of the arrow. We can use trigonometry to calculate this value. Since the arrow is shot at a 30-degree angle, we can use the sine function:
Vy = velocity * sin(angle)
Substituting the values given in the problem:
Vy = 49 m/s * sin(30°)
Now, calculate Vy:
Vy = 49 m/s * 0.5
Vy = 24.5 m/s
We can determine the time it takes for the arrow to reach its maximum height using the equation:
t = Vy / g
where g is the acceleration due to gravity (approximately 9.8 m/s²). Substituting the values:
t = 24.5 m/s / 9.8 m/s²
t ≈ 2.5 s
So, it takes approximately 2.5 seconds for the arrow to reach its maximum height.
Now, let's find the maximum height the arrow reaches. We can use the formula for vertical displacement:
s = Vy * t - 0.5 * g * t²
Substituting the values:
s = 24.5 m/s * 2.5 s - 0.5 * 9.8 m/s² * (2.5 s)²
s = 30.625 m
Therefore, the arrow will reach a maximum height of approximately 30.625 meters.
To find the horizontal distance the arrow travels, we can use the horizontal component of the velocity (Vx) and the time (t) calculated earlier.
First, find the initial horizontal velocity component (Vx) using the cosine function:
Vx = velocity * cos(angle)
Substituting the values given in the problem:
Vx = 49 m/s * cos(30°)
Now, calculate Vx:
Vx = 49 m/s * 0.866
Vx = 42.434 m/s
The horizontal distance (d) traveled by the arrow can be found using the formula:
d = Vx * t
Substituting the values:
d = 42.434 m/s * 2.5 s
d = 106.085 m
Therefore, the arrow will travel approximately 106.085 meters horizontally.
Finally, the total time the arrow is in the air can be determined by doubling the time it takes to reach the maximum height:
Total time = 2 * t
Total time = 2 * 2.5 s
Total time = 5 s
So, the arrow will be in the air for approximately 5 seconds.
Vertical problem:
Vi = 49 sin 30 = 24.5 m/s
v = Vi - 9.81 t
at top v = 0
0 = 24.5 - 9.81 t
t = 2.50 seconds upward
so time in air = 2*2.5 = 5 seconds
h = Vi t - 4.9 t^2
at t 2.5
h = 24.5*2.5 - 4.9(6.25)
= 61.25 - 30.625
= 30.625 meters high
Horizontal:
u = 49 cos 30 forever or
at least until the ground hits it.
so
range = 49 cos 30 * 5 seconds