A 1.50 kg snowball is fired from a cliff 12.5 m high with an initial velocity of 14.0 m/s, directed 41.0° above the horizontal.
(a) Using energy techniques, rather than techniques of Chapter 4, find the speed of the snowball as it reaches the ground below the cliff.
(b) What is that speed if, instead, the launch angle is 41.0° below the horizontal?
(c) What is that speed if the mass is changed to 2.50 kg?
I will be happy to critique your thinking on this.
To find the speed of the snowball as it reaches the ground below the cliff using energy techniques, we can use the principles of conservation of mechanical energy.
(a) In order to solve this question, we need to identify the initial and final states of the snowball. The initial state is at the top of the cliff and the final state is when it reaches the ground below the cliff.
At the top of the cliff, the snowball has gravitational potential energy (Ug) and kinetic energy (K). The gravitational potential energy can be calculated using the formula Ug = mgh, where m is the mass (1.50 kg), g is the acceleration due to gravity (9.81 m/s^2), and h is the height (12.5 m).
Ug = (1.50 kg)(9.81 m/s^2)(12.5 m) = 184.57 J
The initial kinetic energy can be calculated using the formula K = 1/2 mv^2, where m is the mass (1.50 kg) and v is the initial velocity (14.0 m/s).
K = 1/2 (1.50 kg)(14.0 m/s)^2 = 147 J
Since no work is done on the snowball and there is no loss in energy due to friction or air resistance, the total mechanical energy at the top of the cliff (Ei) will be equal to the total mechanical energy at the bottom of the cliff (Ef).
Ei = Ug + K = Ef
At the bottom of the cliff, the snowball has only kinetic energy. Let's represent the final velocity as vf.
Ef = Kf = 1/2 mvf^2
Since the snowball is at the bottom of the cliff, the height (h) is 0. Therefore, the gravitational potential energy (Ug) is also 0.
Now, let's rewrite the equation using the known values:
Ei = Ef
Ug + K = Kf
184.57 J + 147 J = 1/2 (1.50 kg) vf^2
Now, we can solve for vf:
331.57 J = 0.75 kg vf^2
Divide both sides by 0.75 kg:
vf^2 = (331.57 J) / (0.75 kg)
vf^2 = 442.09 m^2/s^2
Take the square root of both sides to find vf:
vf ≈ 21.02 m/s
Therefore, the speed of the snowball as it reaches the ground below the cliff is approximately 21.02 m/s.
(b) If the launch angle is 41.0° below the horizontal, we can still use the same approach as part (a) to find the speed of the snowball as it reaches the ground. The only difference is that the launch angle affects the initial velocity (vi).
We still calculate the initial velocity (vi) using the same trigonometric approach. However, since the launch angle is below the horizontal, we need to take the negative value of the vertical component of the initial velocity.
vi = 14.0 m/s * sin(41.0°)
Now, we can proceed with the same calculations as part (a), using the calculated initial velocity (vi) instead of the given initial velocity.
(c) If the mass is changed to 2.50 kg, we can use the same calculations as part (a), replacing the given mass (1.50 kg) with the new mass (2.50 kg). This will result in different values for gravitational potential energy (Ug) and kinetic energy (K), but the equation and the approach will remain the same.