Question

There is a person who throws a coin vertically downward with an initial speed of 11.8 m/s from the roof of a building, 34.0 m above the ground. How long does it take the coin to reach the ground? Answer in s.

Answer 1

Answer:

Time taken by the coin to reach the ground is 1.69 s

Given:

Initial speed, v = 11.8 m/s

Height of the building, h = 34.0 m

Solution:

Now, from the third eqn of motion:

$v'^{2} = v^{2} + 2gh$

$v'^{2} = 11.8^{2} + 2\times 9.8\times 34.0 = 805.64$

$v' = \sqrt{805.64} = 28.38 m/s$

Now, time taken by the coin to reach the ground is given by eqn (1):

v' = v + gt

$t = \frac{v' - v}{g} = \frac{28.38 - 11.8}{9.8} = 1.69 s$