An arrow is shot from the parapet of a castle and lands on the plain 35 meters below. If the arrow leaves the bow with a speed of 72 m/s, how fast is it traveling when it strikes the ground?
1/2 m v^2 + m g h = 1/2 m V^2
1/2 m 72^2 + m 9.8 35 = 1/2 m V^2
72^2 + (9.8 * 70) = V^2
Thanks so much :)
To find the speed at which the arrow strikes the ground, we can use the principles of kinematics and the equations of motion.
Let's break down the problem and identify the given information:
- Initial velocity (u) of the arrow: 72 m/s (Given)
- Vertical displacement (s) of the arrow: 35 meters (Given)
- Final velocity (v) of the arrow: ? (To be determined)
The motion of the arrow can be divided into two parts:
1. Vertical motion while the arrow is in the air.
2. Horizontal motion, which we assume does not affect the vertical speed.
First, let's find the time taken (t) for the arrow to reach the ground. To do this, we need to consider the vertical motion of the arrow. We can use the second equation of motion:
s = ut + (1/2)at^2
Here, s is the vertical displacement, u is the initial velocity, a is the acceleration (due to gravity, approximately -9.8 m/s^2), and t is the time.
Substituting the given values:
35 = 0t + (1/2)(-9.8)(t^2)
Rearranging the equation:
4.9t^2 = 35
Dividing throughout by 4.9:
t^2 = 35 / 4.9
t^2 ≈ 7.1429
Taking the square root of both sides:
t ≈ √7.1429
t ≈ 2.6743 seconds
Now, we have the time it takes for the arrow to reach the ground.
To find the final velocity (v) of the arrow when it strikes the ground, we can use the first equation of motion:
v = u + at
Here, u is the initial velocity, a is the acceleration (due to gravity), and t is the time.
Substituting the given values:
v = 72 + (-9.8)(2.6743)
Calculating:
v ≈ 72 - 26.151
v ≈ 45.849 ≈ 45.85 m/s
Therefore, the speed of the arrow when it strikes the ground is approximately 45.85 m/s.