Question

1. formula 1 point a box slides with a velocity of 3.1 m/s across a smooth table before falling off. if the box lands 1.9 meters from the table, then how long was the box in the air for? round to 2 decimal places. answer 0.61 6. formula 1 point a student throws a tennis ball straight up and catches it at the same height on the way down. if the ball is in the air for 3.2 seconds, then what was the initial velocity of the ball? answer 7. formula 1 point a ball is thrown straight up with an initial velocity of 38 m/s. how high will it go? round to 1 decimal place. answer

Explanation:

Step1: Identify the relevant kinematic - equation for vertical motion

For vertical - motion problems, we can use the kinematic equation \(v^{2}=v_{0}^{2}-2gh\). At the maximum height, the final velocity \(v = 0\). The acceleration due to gravity \(g = 9.8\ m/s^{2}\) and the initial velocity \(v_{0}=38\ m/s\).

Step2: Rearrange the equation to solve for height \(h\)

Starting with \(v^{2}=v_{0}^{2}-2gh\), when \(v = 0\), we can solve for \(h\):
\[h=\frac{v_{0}^{2}}{2g}\]
Substitute \(v_{0}=38\ m/s\) and \(g = 9.8\ m/s^{2}\) into the formula:
\[h=\frac{38^{2}}{2\times9.8}=\frac{1444}{19.6}\approx73.7\ m\]

Step1: Identify the kinematic - equation for vertical motion

We use the kinematic equation \(y = v_{0}t-\frac{1}{2}gt^{2}\). Since the ball returns to the same height (\(y = 0\)), the equation becomes \(0=v_{0}t-\frac{1}{2}gt^{2}\). We can also use the equation \(v = v_{0}-gt\). When the ball reaches the maximum - height and then comes back to the starting height, the time of flight \(t = 3.2\ s\). The equation \(v = v_{0}-gt\) can be used to find \(v_{0}\) (at the start of the motion). At the end of the motion (when the ball is caught at the same height), if we consider the upward - motion and downward - motion symmetry, we can use the equation \(v = v_{0}-gt\). When the ball is caught at the same height, the displacement \(y = 0\). We can also use the formula \(v = v_{0}-gt\). Rearranging for \(v_{0}\) gives \(v_{0}=v + gt\). At the end of the motion, the velocity has the same magnitude as the initial velocity but opposite direction. If we consider the upward motion only, at the maximum height \(v = 0\). The total time of flight \(t = 3.2\ s\), and the time to reach the maximum height is \(t_{up}=\frac{t}{2}=1.6\ s\). Using \(v = v_{0}-gt\) (where \(v = 0\) at the maximum height), we have \(v_{0}=gt_{up}\). Substituting \(g = 9.8\ m/s^{2}\) and \(t_{up}=1.6\ s\), we get \(v_{0}=9.8\times1.6 = 15.7\ m/s\)

Step1: Identify the horizontal - motion equation

In horizontal motion (assuming no air - resistance), the horizontal velocity \(v_{x}\) is constant. The equation for horizontal motion is \(x = v_{x}t\), where \(x\) is the horizontal displacement, \(v_{x}\) is the horizontal velocity, and \(t\) is the time of flight. We are given \(x = 1.9\ m\) and \(v_{x}=3.1\ m/s\).

Step2: Rearrange the equation to solve for time \(t\)

From \(x = v_{x}t\), we can solve for \(t\) as \(t=\frac{x}{v_{x}}\).
Substitute \(x = 1.9\ m\) and \(v_{x}=3.1\ m/s\) into the formula:
\[t=\frac{1.9}{3.1}\approx0.61\ s\]

Answer:

73.7