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 $2.8 \mathrm { ft }$ above the ground. It leaves the bat with an initial velocity of $149 \mathrm { ft } / \mathrm { sec }$ at a launch angle of $24 ^ { \circ }$. At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of $- 9.7 \mathrm { i } ( \mathrm { ft } / \mathrm { sec } )$ to the ball's initial velocity. Find a vector equation for the path of the baseball.
A) $\mathbf { r } = \left( 149 \cos 24 ^ { \circ } - 9.7 \right) \mathrm { t } + \left( 2.8 + \left( 149 \sin 24 ^ { \circ } \right) \mathrm { t } + 16 \mathrm { t } ^ { 2 } \mathbf { j } \right.$
B) $\mathbf { r } = \left( 149 \cos 24 ^ { \circ } - 9.7 \right) \mathrm { i } + \left( 2.8 + \left( 149 \sin 24 ^ { \circ } \right) \mathrm { t } - 16 \mathrm { t } ^ { 2 } \right) \mathbf { j }$
C) $\mathbf { r } = \left( \left( 149 \cos 24 ^ { \circ } \right) \mathrm { t } - 9.7 \right) \mathbf { i } + \left( 2.8 + \left( 149 \sin 24 ^ { \circ } \right) \mathrm { t } - 16 \mathrm { t } ^ { 2 } \right) \mathbf { j }$
D) $\mathbf { r } = \left( 149 \sin 24 ^ { \circ } - 9.7 \right) \mathrm { t } \mathbf { i } + \left( 2.8 + \left( 149 \cos 24 ^ { \circ } \right) \mathbf { t } - 16 \mathrm { t } ^ { 2 } \right) \mathbf { j }$

Question

Quizplus · Accepted Answer

This problem involves projectile motion with an initial velocity and an added component due to a gust of wind. The x- and y-components of the motion are independent of each other, so we can solve for them separately.

In the x-direction, the ball experiences no acceleration due to air resistance, so its motion is uniform with a constant velocity of $(149\cos 24^\circ - 9.7)\mathrm{ft/s}$. Therefore, the equation for the x-direction is:

$$x = (149\cos 24^\circ - 9.7)t$$

In the y-direction, the ball experiences constant acceleration due to gravity and no horizontal acceleration (since air resistance is neglected). Therefore, we can use the kinematic equation:

$$y = y_0 + v_{0y}t + \frac{1}{2}at^2$$

where $y_0$ is the initial height ($2.8\mathrm{ft}$), $v_{0y}$ is the initial vertical velocity ($149\sin 24^\circ \mathrm{ft/s}$), $a$ is the acceleration due to gravity ($-32\mathrm{ft/s^2}$), and $t$ is the time. Plugging in the values, we get:

$$y = 2.8 + (149\sin 24^\circ)t - 16t^2$$

Therefore, the vector equation for the path of the baseball is:

$$\mathbf{r} = \left((149\cos 24^\circ - 9.7)t\right)\mathbf{i} + \left(2.8 + (149\sin 24^\circ)t - 16t^2\right)\mathbf{j}$$

which matches choice B.

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

Solve the Problem $2.8 \mathrm { ft }$ Above the Ground $149 \mathrm { ft } / \mathrm { sec }$