Answer :
a) Initial energy: 70569 J.
b) Work by air friction at max height: 137979 J.
c) No impact; reaches max height due to negative energy.
a.) Initial total mechanical energy of the projectile:
The initial total mechanical energy (E) of the projectile is the sum of its kinetic energy (KE) and potential energy (PE) when it is at the top of the cliff.
E_initial = KE_initial + PE_initial
The kinetic energy of the projectile at the top of the cliff is zero because it is not moving yet.
KE_initial = 0
The potential energy at the top of the cliff is given by:
PE_initial = m * g * h
Where:
m = mass of the projectile (50.0 kg)
g = acceleration due to gravity (approximately 9.81 m/s^2)
h = height of the cliff (142 m)
PE_initial = 50.0 kg * 9.81 m/s^2 * 142 m = 70569 J
So, the initial total mechanical energy is:
E_initial = 0 + 70569 J = 70569 J
b.) Work done on the projectile by air friction at the maximum height:
At the maximum height, the projectile's speed is 85.0 m/s. We can use the conservation of energy to find the work done by air friction.
E_max_height = KE_max_height + PE_max_height
We already know the initial mechanical energy (E_initial) and the potential energy at the maximum height can be calculated using:
PE_max_height = m * g * h_max
Where:
h_max = maximum height above the ground (427 m)
PE_max_height = 50.0 kg * 9.81 m/s^2 * 427 m = 208548 J
So, the initial kinetic energy (KE_initial) is:
KE_initial = E_initial - PE_initial = 70569 J - 208548 J = -137979 J
Now, the work done by air friction can be calculated as the change in kinetic energy:
Work = ΔKE = KE_max_height - KE_initial = 0 - (-137979 J) = 137979 J
Therefore, the work done by air friction is 137979 J.
c.) In this part, we need to consider the work done by air friction during the entire trajectory. Given that air friction does 1.5 times as much work when the projectile is going down as it did when it was going up, the work done when going down is 1.5 times the work done when going up, which we calculated in part b.
Speed of the projectile before hitting the ground:
Work_done_down = 1.5 * 137979 J = 206968.5 J
Now, we can use the conservation of energy to find the speed of the projectile just before it hits the ground.
E_initial = KE_final + PE_final + Work_done_down
Where:
KE_final = final kinetic energy just before hitting the ground (we'll denote it as K)
PE_final = 0 J (ground level)
E_initial = K + 0 + 206968.5 J
K = E_initial - Work_done_down
K = 70569 J - 206968.5 J = -136399.5 J
To know more about mechanical energy:
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The complete question is- A 50.0-kg projectile is fired at an angle of 30 degrees above the horizontal with an initial speed of 1.20 x 102 m/s from the top of a cliff 142m above level ground, where the ground is taken to be y=0. a.) What is the initial total mechanical energy of the projectile? b.) Suppose the projectile is traveling 85.0 m/s at its maximum height of y=427m. How much work has been done on the projectile by air friction? c.)What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?
