Question

You throw a 0.05-kg golf ball horizontally off a cliff. At the same instance and same height off the ground, your friend throws a 5-kg bowling bowl straight up in the air, reaching a peak height 1 m above the initial throwing height, before it falls down the cliff. Which one hits the ground first?

Answer 1

Answer:

Golf ball

Explanation:

Time of flight when the gold ball is thrown horizontally:

h = 0 + 0.5 gt'²

$t'= \sqrt{\frac{2h}{g}$

When the bowling ball is thrown vertically upwards, the net vertical displacement = h. vertical velocity = u

$h= - ut + 0.5 gt^2 \\ \Rightarrow 0.5gt^2-ut-h = 0\\ \Rightarrow gt^2-2ut-2h =0$

$t = \frac{2u \pm \sqrt{(-2u)^2-4(g)(-2h)}}{2g}\\ t=\frac{2u \pm \sqrt{4u^2+8gh}}{2g}$

$t=\frac{u\pm\sqrt{u^2+2gh}}{g}$

t'<t

The golf ball will hit the ground first.