Assuming no air resistance, how long does it take a penny to fall if thrown down with an initial velocity of 5.0 m/s from the CN tower (553m). Acceleration due to gravity is 9.8 m/s2?

Jun 12, 2015

$t \cong 10.1 s$.

Explanation:

First of all, I suppose that the velocity is a vertical velocity downward.

The motion is an accelerated one with the law:

$s = {s}_{0} + {v}_{0} t + \frac{1}{2} g {t}^{2}$;

so, assuming the origin of the motion up on the tower:

$553 = 0 + 5 t + 0.5 \cdot 9.8 \cdot {t}^{2} \Rightarrow$

$4.9 {t}^{2} + 5 t - 553 = 0$

$\Delta = {b}^{2} - 4 a c = 25 - 4 \cdot 4.9 \cdot \left(- 553\right) \cong {104}^{2}$

$t = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 5 \pm 104}{2 \cdot 4.9}$

${t}_{1} \cong \frac{- 5 - 104}{9.8} < 0$ not acceptable,

${t}_{2} \cong \frac{- 5 + 104}{9.8} \cong 10.1 s$.