An archer needs to hit a bull’s eye on a target at eye level 60.0 ft away. If the archer fires the arrow from eye level with a speed of 47.0 ft/s at an angle of 25 degree above the horizontal, will the arrow hit the target?
convert 47 ft/s to m/s = 14.3m/s
V = 14.3 m/s
Angle = 25 degree
|Vx|=|V|cos(25)
|Vx|=|14.3 cos(25)= 13 m/s
|Vy|=|V|sin(25)
|Vy|=|14.3 sin(25)= 6.04 m/s
t= Vy x 2 /9.8
t= 6.04m/s x 2 /9.8 m/s^2 = 1.23 s
SX = VX * t
then convert it back to feet
so the arrow will miss the target as the arrow travel about 52.3 feet
To determine if the arrow will hit the target, we can use projectile motion equations. The horizontal and vertical components of the arrow's motion can be calculated separately.
First, let's determine the time it takes for the arrow to reach the target. We can use the vertical component of the motion. We know that:
- Initial vertical velocity (Viy) = vertical component of the initial velocity * sin(angle)
- Final vertical velocity (Vfy) = 0 (at the top point)
- Acceleration due to gravity (g) = -32.2 ft/s^2 (assuming downwards is negative)
Using the equation Vfy = Viy + gt, we can solve for time (t):
0 = Viy - 32.2t
Next, let's calculate the horizontal distance the arrow will travel. We know:
- Initial horizontal velocity (Vix) = horizontal component of the initial velocity * cos(angle)
- Distance (d) = Vix * t
Now, we can plug in the given values and calculate the results:
- Vertical component of the initial velocity = 47.0 ft/s * sin(25 degrees)
- Horizontal component of the initial velocity = 47.0 ft/s * cos(25 degrees)
Using these values, we can calculate Viy:
Viy = 47.0 ft/s * sin(25 degrees)
Then we can calculate the time (t):
0 = Viy - 32.2t
Next, let's solve for t:
Viy = 47.0 ft/s * sin(25 degrees)
0 = (47.0 ft/s * sin(25 degrees)) - 32.2t
32.2t = 47.0 ft/s * sin(25 degrees)
t = (47.0 ft/s * sin(25 degrees)) / 32.2
Now that we know the time (t), we can find the horizontal distance (d):
Vix = 47.0 ft/s * cos(25 degrees)
d = Vix * t
Finally, we can check if the horizontal distance is equal to or greater than 60.0 ft (the distance to the target):
If d >= 60.0 ft, the arrow will hit the target.
If d < 60.0 ft, the arrow will not hit the target.
To summarize, calculate the vertical component of the initial velocity (Viy) and the horizontal component of the initial velocity (Vix). Use Viy to calculate the time (t) it takes for the arrow to reach the top point of its trajectory. Then use Vix and t to calculate the horizontal distance (d) the arrow will travel. Compare d to 60.0 ft to determine if the arrow hits the target.
To determine if the arrow will hit the target, we can analyze the horizontal and vertical components of its motion separately.
First, let's calculate the horizontal distance the arrow will travel. This can be found using the horizontal component of the initial velocity:
Vx = V_initial * cos(theta)
Vx = 47.0 ft/s * cos(25 degrees)
Vx ≈ 42.457 ft/s
The time it takes for the arrow to travel 60.0 ft horizontally can be found using the formula:
time = distance / velocity
time = 60.0 ft / 42.457 ft/s
time ≈ 1.413 seconds
Now, let's calculate the vertical distance the arrow will travel. The vertical component of the initial velocity can be found using:
Vy = V_initial * sin(theta)
Vy = 47.0 ft/s * sin(25 degrees)
Vy ≈ 19.957 ft/s
Using this initial vertical velocity, we can calculate the maximum height reached by the arrow. This can be determined using the formula for maximum height in projectile motion:
max_height = (Vy^2) / (2 * g)
max_height = (19.957 ft/s)^2 / (2 * 32.2 ft/s^2)
max_height ≈ 6.165 ft
Considering that the target is at eye level (0 ft), we can see that the maximum height reached by the arrow (6.165 ft) is greater than the height of the target.
Since the arrow will pass above the target's height and the horizontal distance is less than 60.0 ft, the arrow will not hit the target.