Question

chapter 6 motion in two dimensions additional problems
lesson 1 projectile motion
challenge practice

1. you are visiting a friend from elementary school who now lives in a small town. one local amusement is the ice - cream parlor, where stan, the short - order cook, slides his completed ice - cream sundaes down the counter at a constant speed of 2.0 m/s to the servers. (the counter is kept very well polished for this purpose.) if the servers catch the sundaes 7.0 cm from the edge of the counter, how far do they fall from the edge of the counter to the point where the servers catch them?
2. a rock is thrown from a 50.0 - m - high cliff with an initial velocity of 7.0 m/s at an angle of 53.0° above the horizontal. find its velocity when it hits the ground below.

check your progress

1. initial velocity two baseballs are pitched horizontally from the same height but at different speeds. the faster ball crosses home plate within the strike zone, but the slower ball is below the batter’s knees. why do the balls pass the batter at different heights?
2. free - body diagram an ice cube slides without friction across a table at a constant velocity. it slides off the table and lands on the floor. draw free - body and motion diagrams of the ice cube at two points on the table and at two points in the air.
3. projectile motion a tennis ball is thrown out a window 28 m high at an initial velocity of 15.0 m/s and 20.0° below the horizontal. how far does the ball move horizontally before it hits the ground?
4. projectile motion what is the maximum height of a softball thrown at 11.0 m/s and 50.0° above the horizontal?

Explanation:

Step1: Analyze horizontal - motion for ice - cream sundaes

The horizontal speed of the ice - cream sundaes is $v_x = 2.0$ m/s and the horizontal distance is $x=7.0$ cm = $0.07$ m. In horizontal motion (where there is no acceleration, $a_x = 0$), we use the formula $x = v_x t$. Solving for time $t$, we get $t=\frac{x}{v_x}$.
$t=\frac{0.07}{2.0}=0.035$ s

Step2: Analyze vertical - motion for ice - cream sundaes

In vertical motion, the initial vertical velocity $v_{0y}=0$ m/s, the acceleration $a = g=9.8$ m/s², and we use the kinematic equation $y=v_{0y}t+\frac{1}{2}at^{2}$. Since $v_{0y} = 0$ m/s, the equation simplifies to $y=\frac{1}{2}gt^{2}$.
Substitute $t = 0.035$ s and $g = 9.8$ m/s² into the equation:
$y=\frac{1}{2}\times9.8\times(0.035)^{2}$
$y=\frac{1}{2}\times9.8\times0.001225 = 0.0060025\approx6.0$ cm

Answer:

$6.0$ cm