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 determine how high the arrow will go, we need to find the maximum height achieved by the arrow's projectile motion. We can calculate this height using the given information: the launch angle and velocity.
First, let's break down the initial velocity into its vertical and horizontal components. The vertical component of the velocity can be found using trigonometry:
Vertical velocity (Vy) = Velocity (V) * sin(angle)
In this case, the velocity (V) is 49 m/s, and the angle is 30 degrees:
Vy = 49 m/s * sin(30°)
Vy = 49 m/s * 0.5
Vy = 24.5 m/s
Next, we can find the time it takes for the arrow to reach its maximum height. At the highest point, the vertical velocity becomes zero. We can use the following equation to calculate this time:
Vertical velocity (Vy) = Initial vertical velocity (Vyi) + (Gravity acceleration (g) * time (t))
As the arrow reaches its highest point, the vertical velocity becomes zero:
0 = Vyi + (g * t)
Since Vyi is 24.5 m/s and g is 9.8 m/s²:
0 = 24.5 m/s + (9.8 m/s² * t)
Solving for t:
-24.5 m/s = 9.8 m/s² * t
t = -24.5 m/s / 9.8 m/s²
t ≈ -2.5 s
We ignore the negative value because time cannot be negative. Hence, it takes approximately 2.5 seconds for the arrow to reach its maximum height.
Finally, we can find the maximum height (h) using the following formula:
h = Vyi * t + (0.5 * g * t²)
Substituting the values:
h = 24.5 m/s * 2.5 s + (0.5 * 9.8 m/s² * (2.5 s)²)
h = 61.25 m + 30.625 m
h ≈ 91.875 m
Therefore, the arrow will reach approximately 91.875 meters in height.