Question

An arrow is shot horizontally from the top of a building and it lands 200m from the foot of the building after 10s.Assuming air resistance is negligible,calculate the initial velocity of the arrow and the height of the building?
Need an answer urgently please

Answer 1

The initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.

What is velocity?

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,

$\rm h= ut+\frac{1}{2} gt^2 \\\\\ h-\frac{1}{2} gt^2 =ut \\\\ 200 - 0.5 \times(9.81) \times 10^2 = 10 u \\\\ u = - 29.05 \ m/sec$

- 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

$\rm H= ut+\frac{1}{2} gt^2 \\\\\ H = 29.05 \times 10 + 0.5 \times 9.81 \times 10^2 \\\\ H = 781 \ m$

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: brainly.com/question/862972

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