Answer to Question #134340 in Physics for Holly Chapman

Question

A person throws a penny off the roof of a building with an initial velocity of -10m/s (she throws it down). If the roof is 125 m above the ground, determine how fast the penny will be going when it reaches street level

Answer 1

We can find the velocity of the penny when it reaches street level from the kinematic equation (also, we choose the upwards as the positive direction):

$$v_f^2 = v_i^2 + 2ad,$$

here, $v_f$ is the final velocity of the penny when it reaches street level, $v_i = -10 \ ms^{-1}$ is the initial velocity of the penny, $a = g = -9.8 \ ms^{-2}$ is the acceleration due to gravity (since we choose the upwards as the positive direction it will be with the sign minus), $d = -125 \ m$ is the displacement of the penny (or the height of the roof above the ground).

Then, we get:

$$v_f = \sqrt{v_i^2 + 2ad},$$ $$v_f = \sqrt{(-10 \ ms^{-1})^2 + 2 \cdot (-9.8 \ ms^{-2}) \cdot (-125 \ m)} = 50.5 \ ms^{-1}.$$

Answer:

$v_f = 50.5 \ ms^{-1}.$