a 5kg mass is dropped from a height of 30m above the ground. determine the velocity of the mass when it is 18m above the ground.
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Mass is irrelevant if air resistance is ignored
a = -g = -9.81 m/s^2
h = Hi + Vi t + (1/2) a t^2
18 = 30 + 0 t - 4.9 t^2
- 12 = -4.9 t^2
t = 1.56 seconds falling
v = Vi + a t
v = 0 - 9.81 (1.56)
v = - 15.4 m/s
thanks
To determine the velocity of the mass when it is 18m above the ground, we can use the equations of motion.
We can consider the initial height, h₁, as 30m, and the final height, h₂, as 18m. The mass is dropped from rest, so the initial velocity, u, is 0 m/s.
The equation relating the initial height, final height, initial velocity, final velocity, and acceleration due to gravity is:
v² = u² + 2aΔh
In this equation:
- v is the final velocity
- u is the initial velocity (0 m/s)
- a is the acceleration due to gravity (approximated as 9.8 m/s²)
- Δh is the change in height (h₂ - h₁)
Substituting the given values into the equation:
v² = 0² + 2 * 9.8 * (18 - 30)
Simplifying:
v² = 2 * 9.8 * (-12)
v² = -235.2
Since velocity cannot be negative in this context, we can eliminate the negative sign.
v = √235.2
v ≈ 15.33 m/s
Therefore, the velocity of the mass when it is 18m above the ground is approximately 15.33 m/s.
To find the velocity of the mass when it is 18m above the ground, we can use the principles of physics. Let's first calculate the potential energy at a height of 30m and 18m.
The potential energy (PE) of an object at a certain height is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s² on Earth), and h is the height.
PE at a height of 30m:
PE1 = (mass) x (acceleration due to gravity) x (height)
= 5kg x 9.8 m/s² x 30m
= 1470 Joules
PE at a height of 18m:
PE2 = (mass) x (acceleration due to gravity) x (height)
= 5kg x 9.8 m/s² x 18m
= 882 Joules
Next, let's use the principle of conservation of energy, which states that the total energy in a system is constant. The total energy (TE) of the system is the sum of the potential energy and the kinetic energy (KE).
TE = PE + KE
At the highest point (30m), when the object is momentarily at rest, all the potential energy is converted to kinetic energy.
PE1 = KE1
At a height of 18m, the potential energy is partially converted into kinetic energy.
PE2 + KE2 = TE
Since the total energy remains constant, we can equate the potential energy at the highest point to the sum of the potential energy and kinetic energy at 18m.
PE1 = PE2 + KE2
Solving for KE2:
KE2 = PE1 - PE2
= 1470 Joules - 882 Joules
= 588 Joules
Now, we can use the kinetic energy equation:
KE = 0.5mv²
Solving for v (velocity):
v² = (2KE) / m
v² = (2 x 588 Joules) / 5kg
V² = 1176 Joules / 5kg
v² = 235.2 m²/s²
Taking the square root of both sides:
v = √(235.2 m²/s²)
v ≈ 15.34 m/s
Therefore, the velocity of the mass when it is 18m above the ground is approximately 15.34 m/s.