Question

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(d) write the equations for the position of the rock with time, using the coordinates in the figure. (use the following as necessary: t. assume that x and y are in meters and t is in seconds. do not let your answers.)
if the velocity is constant in the x - direction, how is the position along the x - axis related to the initial v0x and the time? m
x =
(e) y = 38.0 - 4.9t²
how long (in s) after being released does the rock strike the ground below the cliff?
2.79 s
(f) with what speed (in m/s) and angle of impact (in degrees clockwise from the +x - axis) does the rock land?
vf = 31.4 m/s
θ =
what is the final x - component of velocity? the final y - component of velocity? how can you use them and trigonometric functions to determine the angle? note a positive angle measured below horizontal (clockwise from +x) is required. ° clockwise from the +x - axis

Explanation:

Step1: Analyze horizontal - motion for part (d)

In the x - direction, since the velocity $v_x$ is constant, the position - time equation is $x = v_{0x}t$, where $v_{0x}$ is the initial x - component of velocity and $t$ is time.

Step2: Analyze vertical - motion for part (e)

The vertical - position equation is $y = y_0+v_{0y}t-\frac{1}{2}gt^2$. When the rock strikes the ground $y = 0$. Given $y_0 = 38.0$ m and $g = 9.8$ m/s², we have $0=38.0 + v_{0y}t-4.9t^2$. This is a quadratic equation $4.9t^2 - v_{0y}t - 38.0=0$. Using the quadratic formula $t=\frac{v_{0y}\pm\sqrt{v_{0y}^2 + 4\times4.9\times38.0}}{9.8}$.

Step3: Analyze velocity components for part (f)

The final x - component of velocity $v_x$ is equal to the initial x - component of velocity $v_{0x}$ (because there is no acceleration in the x - direction, $a_x = 0$). The final y - component of velocity is given by $v_y=v_{0y}-gt$. The speed $v=\sqrt{v_x^2 + v_y^2}$ and the angle $\theta=\tan^{- 1}(\frac{v_y}{v_x})$ (with the appropriate sign convention for the direction).

Answer:

(d) $x = v_{0x}t$
(e) Solve $4.9t^2 - v_{0y}t - 38.0 = 0$ for $t$ using the quadratic formula $t=\frac{v_{0y}\pm\sqrt{v_{0y}^2+744.8}}{9.8}$
(f) $v_x = v_{0x}$, $v_y=v_{0y}-gt$, $v=\sqrt{v_x^2 + v_y^2}$, $\theta=\tan^{-1}(\frac{v_y}{v_x})$ (clock - wise from the + x - axis)