Answer to Question #95444 in Mechanics | Relativity for Francisco Granados

Question

a ball is thrown from a point 1.02 m above the ground. the initial velocity is 18.5 m/s at an angle of 26.0 above the horizontal. find the maximum height of the ball above the ground

Answer 1

$H_{max}= h_{0}+ h_{thrown}$

$h_{thrown}= \frac{V_{0}^2}{2g}\times sin^2\alpha$

Where V_o= 18.5 m/s, alpha= 26 degrees

$h_{thrown} = \frac{18.5^2}{2\times 9.81} \times 0.4384^2= 3.35 m$

$H_{max} =1.02+3.35 = 4.37 m$