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Answer:
[tex]u = 4.56m/s[/tex]
[tex]T = 0.94s[/tex]
Explanation:
See comment for complete question
Given
[tex]h = 106cm[/tex] --- Height
Solving (a): The initial speed
To do this, we make use of the third equation of motion
[tex]v^2=u^2+2ah[/tex]
In this case:
[tex]v = 0m/s[/tex] --- final velocity at the maximum height
[tex]a = -g = -9.8m/s^2[/tex]
[tex]h = 106cm[/tex]
Convert height to metres
[tex]h = 1.06m[/tex]
Substitute these values in [tex]v^2=u^2+2ah[/tex]
[tex]0^2 = u^2 +2 *(-9.8) * 1.06[/tex]
[tex]0^2 = u^2 -20.776[/tex]
Collect Like Terms
[tex]u^2 =20.776[/tex]
Take the positive square root of both sides
[tex]u =\sqrt{20.776[/tex]
[tex]u = 4.55806976691[/tex]
[tex]u = 4.56m/s[/tex] --- approximated
Hence, the initial velocity is 4.56m/s
Solving (b): Time spent in the air.
This will be solved using the first equation of motion.
[tex]v = u + at[/tex]
Where:
[tex]v = 0m/s[/tex] --- final velocity at the maximum height
[tex]a = -g = -9.8m/s^2[/tex]
[tex]u = 4.56m/s[/tex]
So, we have:
[tex]0 = 4.56 - 9.8t[/tex]
Collect Like Terms
[tex]9.8t = 4.56[/tex]
Make t the subject
[tex]t = \frac{4.56}{9.8}[/tex]
[tex]t = 0.46530612244[/tex]
[tex]t = 0.47s[/tex] --- approximated
The above is the time it reaches the maximum height.
The time it stays in the air is:
[tex]T = 2t[/tex]
This gives:
[tex]T = 2*0.47s[/tex]
[tex]T = 0.94s[/tex]