An arrow is shot at a 30 degree angle with the horizontal. It has a velocity of 49m/s. The acceleration of gravity is 9.8 m/s2. How high will the arrow go?
An arrow is shot at 30.0° angle with the horizontal. It has a velocity of 49 m/s.
a. How high will it go?
To find the height the arrow will reach, you can use the following equation of motion:
y = (v₀y² - vₙ²)/(2 * a)
Where:
- y is the vertical displacement or height,
- v₀y is the vertical component of the initial velocity,
- vₙ is the final vertical velocity,
- a is the acceleration due to gravity.
In this case, the arrow is shot at a 30-degree angle with the horizontal, so we need to find the vertical component of the initial velocity (v₀y).
v₀y = v₀ * sin(θ)
Given that the velocity (v₀) is 49 m/s and the angle (θ) is 30 degrees, we can calculate v₀y:
v₀y = 49 m/s * sin(30°)
Next, let's calculate vₙ, which is the final vertical velocity. When the arrow reaches its max height, the vertical velocity becomes zero.
vₙ = 0 m/s
Now we can substitute the values into the equation to find the height (y):
y = (v₀y² - vₙ²)/(2 * a)
= (v₀y² - 0²)/(2 * a)
= v₀y² / (2 * a)
Now we can calculate the height by substituting the values of v₀y and a into the equation:
y = (49 m/s * sin(30°))² / (2 * 9.8 m/s²)
To find the maximum height reached by the arrow, we can use the kinematic equation for vertical motion. The equation is:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity (which is zero at the highest point)
- vi is the initial velocity
- a is the acceleration
- d is the displacement or height
In this case, the arrow is shot vertically upwards at an angle of 30 degrees with the horizontal. Therefore, we need to find the vertical component of the initial velocity.
The vertical component of the initial velocity (viy) can be found using trigonometry. It is given by:
viy = vi * sin(theta)
Where:
- vi is the initial velocity of 49 m/s
- theta is the angle of 30 degrees
Plugging these values into the equation, we have:
viy = 49 m/s * sin(30 degrees)
viy = 24.5 m/s
Now, we can plug the initial vertical velocity (viy), final velocity (vf = 0 m/s), and the acceleration (a = -9.8 m/s^2) into the kinematic equation:
0 = (24.5 m/s)^2 + 2 * (-9.8 m/s^2) * d
Simplifying the equation gives us:
0 = 600.25 m^2/s^2 - 19.6 m/s^2 * d
Re-arranging the equation to solve for the displacement (d), we have:
d = (600.25 m^2/s^2) / (19.6 m/s^2)
d ≈ 30.61 m
Therefore, the arrow will reach a maximum height of approximately 30.61 meters.
Since we don't care about the horizontal motion, just ignore it. The initial vertical velocity
v0 = 49 sin30 = 49 * 1/2 = 24.5
Now, at any time later
v = v0 - at
s = 0 + v0*t - 1/2 a*t^2
= 24.5t - 4.9t^2
= 4.9t(5-t)
So, the arrow is at height=0 at time t=0 and t=5. Therefore, it is at max height when t=2.5
s(2.5) = 24.5*2.5 - 4.9*(2.5)^2 = 30.625 m