The velocity of the ball when it strikes the ground, given the data is 21.56 m/s
<h3>Data obtained from the question</h3>
From the question given above, the following data were obtained:
- Time to reach ground from maximum height (t) = 2.2 s
- Initial velocity (u) = 0 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Final velocity (v) =?
<h3>How to determine the velocity when the ball strikes the ground</h3>
The velocity of the ball when it strikes the ground can be obtained as illustrated below:
v = u + gt
v = 0 + (9.8 × 2.2)
v = 0 + 21.56
v = 21.56 m/s
Thus, the velocity of the ball when it strikes the ground is 21.56 m/s
Learn more about motion under gravity:
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Answer:

Explanation:
When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

Where
is the length of the string and
the velocity of propagation. Use this expression to find the value of
.

The velocity of propagation is given by the expression:

Where
is the desirable variable of the problem, the linear mass density, and
is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

With the value of the tension and the velocity you can find the mass density:


<span>The primary gases of the atmosphere are _____:
</span>
nitrogen, oxygen, and carbondioxide
A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).
We can calculate the initial velocity of the car (u) using the following kinematic equation.
![v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20u%5E%7B2%7D%20%2B%202as%5C%5C%5C%5Cu%20%3D%20%5Csqrt%5B%5D%7Bv%5E%7B2%7D-2as%7D%20%3D%20%5Csqrt%5B%5D%7B%280m%2Fs%29%5E%7B2%7D-2%28-42.61m%2Fs%5E%7B2%7D%20%29%2810.99m%29%7D%20%3D%2030.60m%2Fs)
A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
Learn more: brainly.com/question/14851168