Question

A student throws a coin vertically downward from the top of a building. The coin leaves thrower's hand witha speed of 150 m/s. How far does i fall in 2.00?

Answer 1

distance traveled by the coin in under constant acceleration is given by

$d = v_i*t + \frac{1}{2}at^2$

here vi = 150 m/s

a = 9.8 m/s^2

t = 2 s

now as per above equation

$d = 150*2 + \frac{1}{2}*9.8*2^2$

$d = 300 + 19.6$

$d = 319.6 m$

so the coin will cover 319.6 m distance