Question

a student throws a coin vertically downward from the top of a building. The coin leaves the thrower's hand with a speed of 15.0 m/s. How far does it fall in 2.00s?

Answer 1

Answer: 39.08664m

Normal free fall:
2s = 19.6133m/s
1s = 9.08665m/s

Assuming there's no air resistance:
The first second of the fall it will travel 15m
since the initial speed of the ball is 15m/s

this leaves us with 1 second, however by the time the ball ha traveled 1 second it will be accelerated by the force of gravity (9.08665)

15 + 9.08665 = 24.08665
We're already travelled 15m
and now 24.08665 will be covered in
the last second.

24.08665 + 15 = 39.08664m

-/
if we're gonna be technical:
a coins terminal velocity is only roughly around 19km/s so it wouldn't even travel much

(a brainliest would be appreciated)

Answer 2

Answer:

the coin will move a distance of 30 m

Explanation:

Here the speed of the coin in horizontal direction is given as

$v_x = 15 m/s$

now we know that the time taken by the coin is given as

$t = 2.00 s$

so here we can say that the distance moved by the coin is given as

$d = v_x \times t$

so here we have

$d = (15)(2.00)$

$d = 30 m$

so the coin will move a distance of 30 m