Ke at start = .5 (.125) 28^2
= 49 Joules
.7 * 49 = 34.3 Joules
= m g h = .125 * 9.81 * h
so
h = 28 meters
initial speed of 28 m/s. Assuming 30% loss in total
mechanical energy while ascending, what maximum
height above the position from which it was shot does
the arrow reach?
= 49 Joules
.7 * 49 = 34.3 Joules
= m g h = .125 * 9.81 * h
so
h = 28 meters
The total mechanical energy (E) of the arrow can be calculated using the following equation:
E = Kinetic energy + Potential energy
The kinetic energy (KE) of the arrow is given by the equation:
KE = (1/2) * mass * (velocity^2)
Given:
Mass (m) = 125 g = 0.125 kg
Initial velocity (v) = 28 m/s
Plugging in these values, we can calculate the kinetic energy:
KE = (1/2) * 0.125 * (28^2)
KE ≈ 98 Joules
Next, we need to consider the 30% loss in mechanical energy while ascending. This means that only 70% of the initial mechanical energy will be available at the maximum height. So, the available energy (E_avail) is given by:
E_avail = (70/100) * E
Substituting the value of E, we have:
E_avail = (70/100) * 98
E_avail ≈ 68.6 Joules
Now, we can equate the potential energy (PE) to the available mechanical energy to find the maximum height (h):
PE = E_avail
m * g * h = 68.6
The mass of the arrow (m) and the acceleration due to gravity (g) remain constant, so we can substitute their values:
0.125 * 9.8 * h = 68.6
h ≈ 55.4 meters
Therefore, the maximum height the arrow reaches above the position from which it was shot is approximately 55.4 meters.