Question

Projectile A is dropped from 50 meter building. Projectile B is dropped from another building and takes 2.43 seconds to impact the ground. Upon impact, determine how much faster the right projectile was moving than the slower projectile.

Answer 1

$mgh=mv^2/2\to v=\sqrt{2gh}=\sqrt{2\cdot9.8\cdot50}=31.3(m/s)$

$u=\sqrt{2gh}=\sqrt{2ggt^2/2}=gt=9.8\cdot2.43=23.8\ (m/s)$

$v-u=31.3-23.8=7.5\ (m/s)$ . Answer