The initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.
The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is;
u is the initial velocity of fall = ? m/sec
h is the distance of fall = 200 m
g is the acceleration of free fall = 9.81 m/sec²
H is the height of the building
t is the time period = 10 second
According to Newton's second equation of motion,
[tex]\rm h= ut+\frac{1}{2} gt^2 \\\\\ h-\frac{1}{2} gt^2 =ut \\\\ 200 - 0.5 \times(9.81) \times 10^2 = 10 u \\\\ u = - 29.05 \ m/sec[/tex]
- ve shows the direction is downward.The magnitude of the initial velocity is found as;
u = 29.05 m/sec
The height of the building
[tex]\rm H= ut+\frac{1}{2} gt^2 \\\\\ H = 29.05 \times 10 + 0.5 \times 9.81 \times 10^2 \\\\ H = 781 \ m[/tex]
Hence the initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.
To learn more about the velocity, refer to the link: https://brainly.com/question/862972
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