Question

A projectile fired at 5 m/s from a gun. If the horizontal component of the initial velocity is equal to 4 m/s. What is the value of the vertical component?

Answer 1

Answer:

$the \: value \: of \: the \: vertical \: component \to : \\ (v_v) = 3m {s}^{ - 1}$

Step-by-step explanation:

$let \: the \: velocity \: be \to \: v \\let \: the \: horizontal \: component \: be \to \: v_h \\ let \: the \: vertical \: component \: be \to \: v_v \\ applying \: the \: pythagorean \: rule \: of \: vectors \to : \\ {v}^{2} = {v_v}^{2} + {v_h}^{2} \\ {v_v}^{2} = {v}^{2} - {v_h}^{2} \\ {v_v}^{2} = {5}^{2} - {4}^{2} \\ v_v = \sqrt{(25 - 16)} \\ v_v = \sqrt{9} \\ \boxed{ v_v = 3 {ms}^{ - 1} }$