Question

1. a bowling ball was dropped from the same height and at the same time that a softball was thrown horizontally. neglecting air resistance, which of the statements is true?

a. the bowling ball will hit the ground first
b. the softball will hit the ground first
c. the bowling ball and softball hit the ground at the same time
d. we must know the masses of the balls to determine the time

1. a cannon is placed on top of a cliff that is 20 meters tall. the cannon is fired horizontally with an initial velocity of 45 meters per second. calculate the range of the cannon - ball in meters.

a. 64.3 m
b. 73.2 m
c. 90.9 m
d. 34.8 m
e. 95.8 m

1. a pebble is thrown from ground level with an initial velocity equal to 7 meters per second at an angle of 60 degrees above the horizontal. how long will it take to hit the ground again?

a. 1.15 s
b. 0.62 s
c. 4.25 s
d. 1.24 s
e. 2.02 s

1. a student kicks a soccer ball into the air. at the top of the balls arc what must be true about the y - component of the balls velocity and the balls acceleration? select all true statements.

a. acceleration in the y direction is equal to - 9.8 m/s²
b. acceleration in the y direction is equal to 0 m/s²
c. acceleration in the y direction is equal to 9.8 m/s²
d. v_y = 0 m/s
e. v_y=-9.8 m/s

1. a basketball is thrown with an initial velocity of 10 meters per second at an angle of 45 degrees above the horizontal. the basketball is thrown \granny style\ from ground level and lands back on the ground some distance away from where it was thrown. what is the range of the basketball?

a. 10.2 m
b. 14.4 m
c. 7.1 m
d. 4.5 m
e. 5.1 m

1. a golf cart moves at a constant speed in a straight line. is the golf cart in equilibrium?

a) yes it is in equilibrium
b) no it is not in equilibrium

1. a golf cart moves at a constant speed in a straight line. what must be the net force on the golf cart?

a) the net force is equal to zero
b) the net force is non - zero

Explanation:

Step1: Analyze vertical - motion independence

In projectile motion, the vertical and horizontal motions are independent. For the bowling ball and softball in question 26, the vertical - motion equations are the same since they start from the same height, have the same initial vertical velocity ($v_{0y}=0$ for the dropped bowling ball and also the initial vertical velocity component of the horizontally - thrown softball is 0 in the vertical direction), and experience the same acceleration due to gravity ($g = 9.8\ m/s^{2}$). So they will hit the ground at the same time.

Step2: Solve for time of flight in question 27

The cannon is fired horizontally from a height $h = 20\ m$. In the vertical direction, the initial vertical velocity $v_{0y}=0\ m/s$, and the displacement $y=-h=- 20\ m$, and the acceleration $a=-g=-9.8\ m/s^{2}$. Using the equation $y = v_{0y}t+\frac{1}{2}at^{2}$, we get $-20=0\times t-\frac{1}{2}\times9.8t^{2}$. Solving for $t$:
\[

$$\begin{align*} 4.9t^{2}&=20\\ t&=\sqrt{\frac{20}{4.9}}\approx2.02\ s \end{align*}$$

\]
In the horizontal direction, $v_{0x}=45\ m/s$, and the range $R = v_{0x}t$. So $R = 45\times2.02 = 90.9\ m$.

Step3: Solve for time of flight in question 28

The initial velocity of the pebble is $v_{0}=7\ m/s$ and the launch angle $\theta = 60^{\circ}$. The initial vertical velocity $v_{0y}=v_{0}\sin\theta=7\sin60^{\circ}=7\times\frac{\sqrt{3}}{2}\approx6.06\ m/s$. Using the equation $y = v_{0y}t+\frac{1}{2}at^{2}$, when the pebble hits the ground $y = 0$. So $0 = v_{0y}t-\frac{1}{2}gt^{2}=t(v_{0y}-\frac{1}{2}gt)$. One solution is $t = 0$ (corresponds to the launch time), and the other is $t=\frac{2v_{0y}}{g}=\frac{2\times6.06}{9.8}\approx1.24\ s$.

Step4: Analyze velocity and acceleration at the top of the arc in question 29

At the top of the ball's arc in vertical - projectile motion, the vertical component of the velocity $v_{y}=0\ m/s$ because the ball stops moving upward and is about to start moving downward. The acceleration in the vertical direction is always $a=-g=-9.8\ m/s^{2}$ due to the force of gravity.

Step5: Calculate the range in question 30

The initial velocity $v_{0}=10\ m/s$ and the launch angle $\theta = 45^{\circ}$. The initial vertical velocity $v_{0y}=v_{0}\sin\theta$ and the initial horizontal velocity $v_{0x}=v_{0}\cos\theta$. The time of flight $t$ is given by $y = v_{0y}t-\frac{1}{2}gt^{2}$, and when $y = 0$ (returns to the same height), $t=\frac{2v_{0y}}{g}=\frac{2v_{0}\sin\theta}{g}$. The range $R = v_{0x}t=v_{0}\cos\theta\times\frac{2v_{0}\sin\theta}{g}=\frac{v_{0}^{2}\sin2\theta}{g}$. Since $\theta = 45^{\circ}$, $\sin2\theta=\sin90^{\circ}=1$, and $v_{0}=10\ m/s$, $g = 9.8\ m/s^{2}$. So $R=\frac{10^{2}\times1}{9.8}\approx10.2\ m$.

Step6: Analyze equilibrium in question 31

An object is in equilibrium if the net force acting on it is zero. A golf - cart moving at a constant speed in a straight line has zero acceleration ($a = 0$). According to Newton's second law $F_{net}=ma$, if $a = 0$, then $F_{net}=0$. So the golf - cart is in equilibrium.

Step7: Determine net force in question 32

Since the golf - cart is moving at a constant speed in a straight line, its acceleration $a = 0$. According to Newton's second law $F_{net}=ma$, when $a = 0$, $F_{net}=0$.

Answer:

1. c. the bowling ball and softball hit the ground at the same time
2. c. 90.9 m
3. d. 1.24 s
4. a. Acceleration in the y direction is equal to - 9.8 m/s²; d. $V_{y}=0\ m/s$
5. a. 10.2 m
6. A) Yes it is in equilibrium
7. A) The net force is equal to zero