How high will an arrow travel into the air when shot upwards?
Bow stress: 68 J
Arrow: 26 g
(there is no air resistance)
Key is precious so answer John's question (Ronald won't help you). I need answer too :P
someone help him out, we all need a chance
262
E = mgh
Solve for h
To determine how high an arrow will travel when shot upwards, we can use the principle of conservation of mechanical energy. This principle states that the total mechanical energy of an object remains constant in the absence of external forces (such as air resistance). The mechanical energy is the sum of the arrow's kinetic energy (KE) and its potential energy (PE).
First, let's calculate the initial kinetic energy of the arrow. The kinetic energy (KE) is given by the equation:
KE = (1/2) * mass * velocity^2
Since the arrow is shot upwards, its initial velocity is 0 (at the peak of its trajectory) and its kinetic energy will also be zero.
Next, let's determine the potential energy of the arrow at the peak of its trajectory. The potential energy (PE) is given by the equation:
PE = mass * acceleration due to gravity * height
Since the arrow is at its peak, its velocity is zero, and the entire initial kinetic energy has been converted to potential energy.
To find the height (h) to which the arrow will travel, we can rearrange the equation for potential energy:
height (h) = PE / (mass * acceleration due to gravity)
The mass of the arrow is given as 26 grams, which can be converted to kilograms by dividing by 1000 (1 gram = 0.001 kilograms). Therefore, the mass of the arrow is 0.026 kilograms.
The acceleration due to gravity is approximately 9.8 m/s^2.
Now we can calculate the height:
height (h) = PE / (mass * acceleration due to gravity)
= (mass * acceleration due to gravity * height) / (mass * acceleration due to gravity)
= height
Since the arrow is shot straight upwards, it will come to a stop at the peak of its trajectory before falling back down. At this point, its potential energy is at its maximum, and its kinetic energy is zero. Therefore, all of its initial mechanical energy (0.068 J) is converted to potential energy.
Now we can substitute the values to find the height:
height (h) = 0.068 J / (0.026 kg * 9.8 m/s^2)
= 0.260 m (rounded to three decimal places)
Therefore, the arrow will travel approximately 0.260 meters (or 26 centimeters) into the air when shot upwards.
Please note that this calculation assumes no air resistance and that the arrow is shot in a vacuum. In reality, factors such as air resistance and wind will affect the arrow's trajectory, resulting in a lower maximum height.