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
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The energy becomes 0.50 times in 6.72 s.
Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.
Et=0.50E
1
k(4₂)² = (0.5) - kA²
(4₂)² = (0.5) A²
At = 0.71A
So, the amplitude of the oscillator becomes 0.71 times its initial ar
0.71A = = A(0.96)¹2
log(0.71)
log(0.96)
8.4
n=
So, the time taken for n oscillation is obtained as,
t = n (0.800 s)
= (8.4) (0.800)
= 6.72 s
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