# Question #d3c9f

##### 1 Answer
Apr 10, 2015

$32.1 \text{ m/s }$$\left[{82}^{o} \text{ below horizontal}\right]$

First divide the initial velocity vector into horizontal components and vertical components.

The horizontal component will be

${v}_{x} = v \cos \theta = 7.0 \cos 53$ m/s

I am assuming we are neglecting air resistance so this value will remain constant. ${v}_{f x} = {v}_{i x} = {v}_{x} = 4.21$ m/s

The vertical component will be

${v}_{y} = v \sin \theta = 7.0 \sin 53$ m/s [up]

${v}_{f y}$ the final velocity in the y-direction can be determined by

${\left({v}_{f y}\right)}^{2} = {\left({v}_{i y}\right)}^{2} + 2 g \Delta y$ . . . . substitution gives

${v}_{f y} = \sqrt{{\left(7.0 \sin 53\right)}^{2} + 2 \left(- 9.81\right) \left(- 50\right)} = 31.82$ m/s

Now combine the two components to get the final velocity ${v}_{f} = \sqrt{{v}_{f x}^{2} + {v}_{f y}^{2}} = 32.1$ m/s

$\theta = {\tan}^{- 1} \left(\left({v}_{\text{fy")/(v_"fx}}\right)\right) = {\tan}^{- 1} \left(\frac{31.82}{4.21}\right) = {82}^{o}$