Answer to Question #150351 in Mechanics | Relativity for Caleb Schultz

Question

How far will a jumper go if he accelerates to 11.0 m/sec and jumps at an angle of 30

Answer 1

Horizontal components

$Horizontal\ components \\ V_{ix} = V_{i}\times \cos \Theta \\ V_{ix} = 11.0 m/s \times \cos \ 30\\ V_{ix} = 9.5 m/s \\ Vertical\ components \\ V_{iy} = V_{i}\times \sin \Theta \\ V_{iy} = 11.0 m/s \times \sin \ 30\\ \\ V_{iy} = 6 m/s \\ To\ find\ time \ t \\ V_{fy} = V _{iy}+ a_{y}t\\ Subtituting \ the \ values\ in\ the\ above\ equation\\ -6m/s = 6m/s +(-9.8m/s^2)\\ Solving\ the\ equation\ we\ get\ time\ (t)\\ we\ get \\ t=1.22s\\ To\ get\ how\ far\ will\ the\ jumper\ go (X)\\ \\ X=V_{ix}t +\frac{1}{2}a_{x}t^{2}\\ X=9.5m/s\times1.22s+\frac{1}{2}(0m/s^2)\times(1.22)^2\\ X=11.59m$