Question

1. jeff hits a baseball straight up in the air with an initial velocity of 38.0 m/s. how much does it stay in the air? find total time. 10. how high does the ball in #9 above travel? 11. justin, the famous race car driver, is traveling at an initial velocity of 8.00 m/s and then accelerates uniformly at 10.0 m/s² for 5.00 seconds. how far does he travel in this time interval? 12. what is his final velocity in #11 above? 13. alexis throws a metal ball downward with a velocity of 20.0 m/s from a tall building. if the height of the building is 500 m, with what velocity does the ball hit the ground? 14. an electron in a vacuum tube is decelerated by a charged grid at a rate of -7.00×10⁵ m/s². it decelerates for an interval of 6.00×10⁻³ seconds. the speed of the electron after the deceleration takes place is 2.00×10⁵ m/s. what is the speed of the electron before the deceleration? (this is an easy one!) 15. alex is riding on a hot air balloon and is ascending at a rate of 24.0 m/s. he accidentally knocks his bologna sandwich out of the balloon while at an altitude of 200 m. calculate the total time the bologna sandwich was in the air. neglect air resistance. bonus. vickia drops a heavy rock from a cliff 40 m above the water. it hits the water with a certain velocity and continues to sink to the bottom of the lake at this constant velocity. it reaches the bottom of the lake 12 seconds after it was dropped. how deep is the lake?

Explanation:

Step1: Identify the relevant kinematic equation for vertical - motion

For an object in free - fall, when the initial velocity is \(v_0\) and the acceleration \(a=-g=- 9.8\ m/s^{2}\) (taking up as positive and the acceleration due to gravity acts downwards), and the displacement \(y - y_0 = 0\) (returns to the same height), the equation \(y - y_0=v_0t+\frac{1}{2}at^{2}\) is used.

Step2: Substitute the values into the equation

Given \(v_0 = 38.0\ m/s\), \(a=-9.8\ m/s^{2}\), and \(y - y_0 = 0\), we have \(0 = 38.0t-\frac{1}{2}\times9.8t^{2}\). Factor out \(t\): \(t(38.0 - 4.9t)=0\). One solution is \(t = 0\) (corresponds to the initial time). The other non - zero solution is obtained by solving \(38.0-4.9t = 0\).

Step3: Solve for \(t\)

\[

$$\begin{align*} 4.9t&=38.0\\ t&=\frac{38.0}{4.9}\approx7.76\ s \end{align*}$$

\]

Answer:

The ball stays in the air for approximately \(7.76\ s\)