Answer: Height of building = 17.69m, velocity of brick = 18.6m/s
Explanation: From the question, the body has a zero initial speed, thus initial velocity (u) all through the motion is zero.
By ignoring air resistance makes it a free fall motion thus making it to accelerate constantly with a value of [tex]a = 9.8m/s^{2}[/tex].
Time taken to fall = 1.90s
a)
thus the height of the building is calculated using the formulae below
[tex]H = ut + \frac{1}{2} gt^{2}[/tex]
but u = 0 , hence
[tex]H = \frac{1}{2} gt^{2}[/tex]
[tex]H = \frac{1}{2} *9.8* 1.9^{2} \\\\H = 17.69m[/tex]
b)
to get the value of velocity (v) as the brick hits the ground, we use the formulae below
[tex]v^{2} = u^{2} + 2aH[/tex]
but u= 0, hence
[tex]v^{2} = 2gH\\[/tex]
[tex]v^{2} = 2 * 9.8 * 17.69\\v = \sqrt{2 *9.8* 17.69} \\v = 18.62m/s[/tex]
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find the attachment in this answer for the accleration- time graph, velocity- time graph and distance time graph
a. The height of building, in meters, is equal to 17.69 meters.
b. The magnitude of the brick’s velocity just before it reaches the ground is equal to 18.72 m/s.
Given the following data:
We know that acceleration due to gravity (a) for an object in free fall is equal to 9.8 meter per seconds square.
a. To determine the height of building, in meters, we would use the second equation of motion:
Mathematically, the second equation of motion is given by the formula;
[tex]S = ut + \frac{1}{2} at^2[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]S = 0(1.90) + \frac{1}{2} \times 9.8 \times 1.90^2\\\\S = 0 + 4.9 \times 3.61[/tex]
Distance, S = 17.69 meters.
b. To determine the magnitude of the brick’s velocity just before it reaches the ground, we would use the third equation of motion;
[tex]V^2 = U^2 + 2aS[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]V^2 = 0^2 + 2 \times 9.8 \times 17.69\\\\V^2 = 350.26\\\\V = \sqrt{350.26}[/tex]
V = 18.72 m/s
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