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.
"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
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