A 125-g arrow is shot vertically upward with an
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?
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
To find the maximum height the arrow reaches, we first need to calculate the total mechanical energy of the arrow when it was shot and then subtract the loss due to air resistance.
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.