Question

example problem 2 - the flight of a ball

1. a player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the horizontal. find each of the following. assume that forces from the air on the ball are negligible.

a. the balls hang time
b. the balls maximum height
c. the horizontal distance the ball travels before hitting the ground

1. the player in the previous problem then kicks the ball with the same speed but at 60.0° from the horizontal. what is the balls hang time, horizontal distance traveled, and maximum height?
2. lucy and her friend are working at an assembly plant making wooden toy giraffes. at the end of the line, the giraffes go horizontally off the edge of a conveyor belt and fall into a box below. if the box is 0.60 m below the level of the conveyor belt and 0.40 m away from it, what must be the horizontal velocity of giraffes as they leave the conveyor belt?
3. the horizontal and vertical components of the stones velocity just before it hits the ground? the horizontal component of velocity is 5.0 m/s and the vertical component is 20 m/s. how far from the base of the cliff does the stone hit the ground?

Explanation:

Step1: Analyze vertical - motion for hang time

The initial vertical velocity is $v_{0y}=v_0\sin\theta$, where $v_0 = 27.0$ m/s and $\theta = 30.0^{\circ}$. So $v_{0y}=27.0\sin30.0^{\circ}=27.0\times0.5 = 13.5$ m/s. Using the kinematic equation $v = v_0+at$ for vertical motion, at the highest - point $v_y = 0$, and $a=-g=- 9.8$ m/s². The time to reach the highest - point $t_1=\frac{v_y - v_{0y}}{a}=\frac{0 - 13.5}{-9.8}\approx1.38$ s. The hang time $T = 2t_1=2\times1.38 = 2.76$ s.

Step2: Analyze vertical - motion for maximum height

Using the kinematic equation $v_y^{2}-v_{0y}^{2}=2a\Delta y$. At the maximum height $v_y = 0$, $v_{0y}=13.5$ m/s and $a=-9.8$ m/s². So $\Delta y=\frac{v_y^{2}-v_{0y}^{2}}{2a}=\frac{0-(13.5)^{2}}{2\times(-9.8)}=\frac{- 182.25}{-19.6}\approx9.30$ m.

Step3: Analyze horizontal motion for horizontal distance

The initial horizontal velocity $v_{0x}=v_0\cos\theta$, where $v_0 = 27.0$ m/s and $\theta = 30.0^{\circ}$, so $v_{0x}=27.0\cos30.0^{\circ}=27.0\times\frac{\sqrt{3}}{2}\approx23.4$ m/s. Using the equation $x = v_{0x}T$, with $T = 2.76$ s and $v_{0x}\approx23.4$ m/s, we get $x=23.4\times2.76\approx64.6$ m.

Answer:

a. Hang time: $T = 2.76$ s
b. Maximum height: $h\approx9.30$ m
c. Horizontal distance: $x\approx64.6$ m