A 1.50 kg snowball is fired from a cliff 13.0 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.
m/s
(b) What is that speed if, instead, the launch angle is 41.0° below the horizontal?
m/s
(c) What is that speed if the mass is changed to 2.50 kg
m/s
Please show your work. Someone will ne glad to assist you.
86
To find the speed of the snowball as it reaches the ground below the cliff, we can use the principle of conservation of energy. The initial potential energy of the snowball is converted into kinetic energy as it falls.
Let's go step by step to find the answers to each question:
(a) Using energy techniques:
We need to calculate the potential energy and the kinetic energy of the snowball at the top and the bottom of the cliff.
At the top of the cliff:
Potential energy (PE) = mgh
PE = (1.50 kg)(9.8 m/s^2)(13.0 m) = 191.1 J
The initial kinetic energy (KE) is zero since the snowball is not moving horizontally at the top of the cliff.
At the bottom of the cliff:
The potential energy is zero since the snowball is at ground level.
The final kinetic energy is equal to the initial potential energy:
KE = PE = 191.1 J
To find the final velocity (v), we can use the formula for kinetic energy:
KE = (1/2)mv^2
Rearranging the formula:
v^2 = (2KE)/m
v^2 = (2)(191.1 J) / 1.50 kg
v^2 = 255.2 m^2/s^2
Taking the square root of both sides:
v = √(255.2 m^2/s^2)
v ≈ 15.974 m/s
Therefore, the speed of the snowball as it reaches the ground below the cliff is approximately 15.974 m/s.
(b) Repeat the same calculations, but with the launch angle of 41.0° below horizontal:
The only difference is that now the snowball is launched downward. Therefore, we need to consider the launch angle as -41.0° in our calculations.
The potential energy at the top of the cliff remains the same.
At the bottom of the cliff, the kinetic energy remains the same as in part (a) since the speed and mass are the same.
Performing the same calculations, we find:
v = √(255.2 m^2/s^2)
v ≈ 15.974 m/s
Therefore, the speed of the snowball as it reaches the ground below the cliff is the same, approximately 15.974 m/s, regardless of the launch angle.
(c) Now, let's consider a snowball with a mass of 2.50 kg:
Repeat the same calculations with a different mass:
Potential energy at the top of the cliff remains the same.
At the bottom of the cliff, we need to recalculate the final kinetic energy using the new mass:
KE = (1/2)mv^2
KE = (1/2)(2.50 kg)(v^2)
Setting KE equal to the initial potential energy:
KE = (1/2)(2.50 kg)(v^2) = 191.1 J
Rearranging the formula:
v^2 = (191.1 J)(2) / (2.50 kg)
v^2 = 152.88 m^2/s^2
Taking the square root of both sides:
v = √(152.88 m^2/s^2)
v ≈ 12.370 m/s
Therefore, the speed of the snowball as it reaches the ground below the cliff, with a mass of 2.50 kg, is approximately 12.370 m/s.