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?
To find the height the arrow will reach, we can use the equations of motion in projectile motion.
The horizontal motion and vertical motion of the arrow are independent of each other.
First, let's break down the given information:
- The velocity of the arrow is 49 m/s.
- The angle with the horizontal is 30 degrees.
- The acceleration due to gravity is 9.8 m/s^2.
The vertical motion can be analyzed using the following equation:
Projectile motion equation:
h = (v^2 * sin^2θ) / (2g)
Where:
h = maximum height
v = initial velocity of the arrow
θ = launch angle with respect to the horizontal
g = acceleration due to gravity
Now, let's plug in the known values into the equation:
h = (49^2 * sin^2(30)) / (2 * 9.8)
To simplify the calculation, we need to find the value of sin^2(30):
sin(30) = 0.5
sin^2(30) = 0.5 * 0.5 = 0.25
Now, we can substitute the values:
h = (49^2 * 0.25) / (2 * 9.8)
h = (2401 * 0.25) / 19.6
h = 600.25 / 19.6
h ≈ 30.64 meters
Therefore, the arrow will reach a maximum height of approximately 30.64 meters.