Suppose the rocket is coming in for a vertical landing at the surface of the earth. The captain adjusts the engine thrust so tha
1 answer:
You might be interested in
Explanation:
Below is an attachment containing the solution.
Answer:4.05 s
Explanation:
Given
First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s
Both hit the ground at the same time
Let h be the height of cliff and it reaches after time t
For second stone
---2
Equating 1 &2 we get
Answer:
h = v₀² / 2g , h = k/4g x²
Explanation:
In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches
Starting point. Lower compressed spring
Em₀ = K = ½ m v²
Final point. Highest on the path
= U = mg h
As or no friction the energy is conserved
Em₀ = Em_{f}
½ m v₀²² = m g h
h = v₀² / 2g
We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse
½ k x² = 2 g h
h = k/4g x²