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Solve the Problem 2.8ft2.8 \mathrm { ft } Above the Ground 149ft/sec149 \mathrm { ft } / \mathrm { sec }

Question 74

Multiple Choice

Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface
-A baseball is hit when it is 2.8ft2.8 \mathrm { ft } above the ground. It leaves the bat with an initial velocity of 149ft/sec149 \mathrm { ft } / \mathrm { sec } at a launch angle of 2424 ^ { \circ } . At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of 9.7i(ft/sec) - 9.7 \mathrm { i } ( \mathrm { ft } / \mathrm { sec } ) to the ball's initial velocity. Find a vector equation for the path of the baseball.


A) r=(149cos249.7) t+(2.8+(149sin24) t+16t2j\mathbf { r } = \left( 149 \cos 24 ^ { \circ } - 9.7 \right) \mathrm { t } + \left( 2.8 + \left( 149 \sin 24 ^ { \circ } \right) \mathrm { t } + 16 \mathrm { t } ^ { 2 } \mathbf { j } \right.
B) r=(149cos249.7) i+(2.8+(149sin24) t16t2) j\mathbf { r } = \left( 149 \cos 24 ^ { \circ } - 9.7 \right) \mathrm { i } + \left( 2.8 + \left( 149 \sin 24 ^ { \circ } \right) \mathrm { t } - 16 \mathrm { t } ^ { 2 } \right) \mathbf { j }
C) r=((149cos24) t9.7) i+(2.8+(149sin24) t16t2) j\mathbf { r } = \left( \left( 149 \cos 24 ^ { \circ } \right) \mathrm { t } - 9.7 \right) \mathbf { i } + \left( 2.8 + \left( 149 \sin 24 ^ { \circ } \right) \mathrm { t } - 16 \mathrm { t } ^ { 2 } \right) \mathbf { j }
D) r=(149sin249.7) ti+(2.8+(149cos24) t16t2) j\mathbf { r } = \left( 149 \sin 24 ^ { \circ } - 9.7 \right) \mathrm { t } \mathbf { i } + \left( 2.8 + \left( 149 \cos 24 ^ { \circ } \right) \mathbf { t } - 16 \mathrm { t } ^ { 2 } \right) \mathbf { j }

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