A girl shoots an arrow from the top of a cliff. The arrow is initially at a point 20 meters above the level field below. The arrow is shot at an angle of 30 degrees above the horizontal with a speed of 39.2m/s. How far out from the base of the cliff will the arrow land?
You need to first find the time in the air, from the vertical flight equation. You can figure the initial vertical velocity from the angle.
hcliff=vi*time -1/2 g time^2
solve for time.
Then, use that time for the horizontal equation..
d=vhorizontal*tiem
i tried that. it doesnt work
nvm it works. thanksssssss
To find out how far out from the base of the cliff the arrow will land, we need to break down the problem into horizontal and vertical components.
First, let's calculate the time it takes for the arrow to reach the ground. We can use the vertical motion equation:
y = y0 + v0yt - 0.5gt^2
Where:
y is the final vertical displacement (0 meters, since the arrow hits the ground)
y0 is the initial vertical displacement (20 meters)
v0y is the vertical component of the initial velocity (v0 * sinθ, where v0 is the initial speed and θ is the angle of projection)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time it takes for the arrow to reach the ground
Since y equals 0, we can rearrange the equation to solve for t:
0 = 20 + (39.2 * sin(30))t - 0.5 * 9.8 * t^2
Simplifying the equation yields:
4.9t^2 - 19.6t - 20 = 0
Solving this quadratic equation, we find that t ≈ 2.4 seconds (taking the positive root).
Now, we can calculate the horizontal distance traveled by the arrow using the horizontal motion equation:
x = x0 + v0xt
Where:
x is the horizontal displacement (what we want to find)
x0 is the initial horizontal displacement (0 meters since the arrow is shot from the top of the cliff)
v0x is the horizontal component of the initial velocity (v0 * cosθ, where v0 is the initial speed and θ is the angle of projection)
t is the time it takes for the arrow to reach the ground (which we found to be approximately 2.4 seconds)
Since x0 equals 0 and v0x equals v0 * cosθ, we can substitute these values into the equation:
x = (39.2 * cos(30)) * 2.4
Simplifying the equation:
x ≈ 67.9 meters
Therefore, the arrow will land approximately 67.9 meters out from the base of the cliff.