Question

golf ball hit w/ velocity of 60.0m/s at angle of 35.0 degree above horizontal. a.) how far ball travels horizontally. b.) find max height the ball reaches vertically. c.) find resultant velocity of velocity after 5.00 s of flight. the ball and direction of velocity: viy 60sin35, vix 60cos35, ay=-9.8, ax = 0, ax = vit+1/2at^2 → 60cos35t + 1/2·0·t^2

Explanation:

Step1: Find initial horizontal and vertical velocities

The initial velocity $v_0 = 60.0$ m/s. The initial horizontal velocity $v_{0x}=v_0\cos\theta$ and initial vertical velocity $v_{0y}=v_0\sin\theta$, where $\theta = 35.0^{\circ}$. So $v_{0x}=60\cos35^{\circ}\approx 49.15$ m/s and $v_{0y}=60\sin35^{\circ}\approx 34.42$ m/s.

Step2: Calculate maximum - height (A)

At maximum - height, the vertical velocity $v_y = 0$. We use the kinematic equation $v_y^2=v_{0y}^2 + 2a_y\Delta y$. Here, $a_y=- 9.8$ m/s². Rearranging for $\Delta y$ gives $\Delta y=\frac{v_y^2 - v_{0y}^2}{2a_y}$. Substituting the values: $\Delta y=\frac{0-(34.42)^2}{2\times(-9.8)}\approx 60.6$ m.

Step3: Calculate horizontal distance (B)

First, we find the time of flight $t$. When the ball returns to the same height, the vertical displacement $\Delta y = 0$. Using the kinematic equation $\Delta y=v_{0y}t+\frac{1}{2}a_y t^2$. Since $\Delta y = 0$, we have $0 = v_{0y}t+\frac{1}{2}a_y t^2=t(v_{0y}+\frac{1}{2}a_y t)$. One solution is $t = 0$ (initial time), and the other is $t=\frac{-2v_{0y}}{a_y}=\frac{-2\times34.42}{-9.8}\approx 7.02$ s. The horizontal distance $x = v_{0x}t$. Substituting $v_{0x}\approx 49.15$ m/s and $t\approx 7.02$ s, we get $x\approx 49.15\times7.02\approx 345$ m.

Step4: Calculate vertical and horizontal velocities at $t = 5.00$ s (C)

The horizontal velocity remains constant: $v_x=v_{0x}\approx 49.15$ m/s. The vertical velocity is given by $v_y=v_{0y}+a_y t$. Substituting $v_{0y}\approx 34.42$ m/s, $a_y=-9.8$ m/s², and $t = 5.00$ s, we get $v_y=34.42-9.8\times5.00=34.42 - 49=-14.58$ m/s. The resultant velocity $v=\sqrt{v_x^2 + v_y^2}=\sqrt{(49.15)^2+(-14.58)^2}\approx 51.2$ m/s. The direction $\theta=\tan^{-1}(\frac{v_y}{v_x})=\tan^{-1}(\frac{-14.58}{49.15})\approx - 16.4^{\circ}$ (below the horizontal).

Answer:

A. Maximum height: approximately 60.6 m
B. Horizontal distance: approximately 345 m
C. Resultant velocity: approximately 51.2 m/s at an angle of approximately $-16.4^{\circ}$ (below the horizontal)