Question

Felix and Oscar are standing on a tower that is 112 feet high. Felix drops his penny while Oscar throws his penny down and it hits the ground in 2 seconds. How long will it take Felix's penny to hit the ground and what was the initial velocity of Oscar's penny?

Answer 1

Answer:

It will take 2.64 s for Felix's penny to hit the ground.

Initial velocity of Oscar's penny is 7.27 m/s or 23.85 ft/s

Step-by-step explanation:

Given:

Height of the tower is, $h=112\ ft$

Felix drops the penny while Oscar throws his penny.

Time taken by Oscar's penny to reach ground is, $t_o=2\ s$

Now, as Oscar throws his penny, there will be some initial velocity of his penny. Let the initial velocity of his penny be $u_o$.

As the penny is falling vertically, acceleration of the penny is due to gravity and is equal to -9.8 m/s².

Now, displacement of the penny is equal to the height of the tower = -112 ft = -34.14 m (1 ft = 0.3048 m)

Using equation of motion,

$y=u_ot+\frac{1}{2}at_o^2\\\\-34.14=2u_o-\frac{1}{2}\times9.8\times 2^2\\\\-34.14=2u_o-19.6\\\\2u_o=-34.14+19.6\\\\2u_o=-14.54\\\\u_o=\frac{-14.54}{2}=-7.27\ m/s$

Now, 1 m/s = 3.28 ft/s

So, 7.27 m/s = $7.27\times 3.28 = 23.85\ ft/s$

Therefore, the initial velocity of Oscar's penny is 7.27 m/s or 23.85 ft/s in the downward direction.

Now, Felix drops the penny. So, initial velocity of Felix's penny is 0 m/s.

Displacement of the penny is same and equal to -112 ft = -34.14 m

Using the same equation of motion,

$y=ut+\frac{1}{2}at^2\\\\-34.14=0-\frac{1}{2}\times9.8\times t^2\\\\-34.14=-4.9t^2\\\\t^2=\frac{-34.14}{-4.9}\\\\t=\sqrt{\frac{-34.14}{-4.9}}\\\\ t=2.64\ s$

Therefore, time taken by Felix's penny to reach ground is 2.64 s.