During a test a rocket travels upward at 90 m/s , and when it is 50 m from the ground its engine fails. Determine the maximum he
ight sb reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m / s 2 due to gravity. Neglect the effect of air resistance.
2 answers:
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Answer:
The velocity of the frozen rock at is -14.711 meters per second.
Explanation:
The frozen rock experiments a free fall, which is a type of uniform accelerated motion due to gravity and air viscosity and earth's rotation effect are neglected. In this case, we need to find the final velocity (), measured in meters per second, of the frozen rock at given instant and whose kinematic formula is:
(Eq. 1)
Where:
- Initial velocity, measured in meters per second.
- Gravity acceleration, measured in meters per square second.
- Time, measured in seconds.
If we get that ,
and
, then final velocity is:
The velocity of the frozen rock at is -14.711 meters per second.
Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m