Question

1. an object dropped from rest will have a velocity of approximately 30. meters per second at the end of a) 1.0 s b) 2.0 s c) 3.0 s d) 4.0 s 14. a ball is thrown straight up with a speed of 12 meters per second near the surface of earth. what is the maximum height reached by the ball? neglect air friction. a) 15 m b) 7.3 m c) 1.2 m d) 0.37 m

Explanation:

Problem 13

Step1: Use the formula for velocity in free - fall

The formula for the velocity of an object in free - fall is \(v = v_0+at\). Since the object is dropped from rest, \(v_0 = 0\) and \(a = g= 10\space m/s^{2}\) (approximate value of acceleration due to gravity near the Earth's surface). So the formula simplifies to \(v = gt\).

Step2: Solve for \(t\)

We know \(v = 30\space m/s\) and \(g = 10\space m/s^{2}\). Substituting into \(v = gt\), we get \(t=\frac{v}{g}\).

$$t=\frac{30\space m/s}{10\space m/s^{2}}=3\space s$$

Step1: Use the kinematic formula \(v^{2}-v_{0}^{2}=2ah\)

At the maximum height, the final velocity \(v = 0\). The initial velocity \(v_0=12\space m/s\) and \(a=-g = - 10\space m/s^{2}\) (negative because it acts against the direction of motion). The formula becomes \(0 - v_{0}^{2}=2(-g)h\).

Step2: Solve for \(h\)

From \(0 - v_{0}^{2}=2(-g)h\), we can express \(h\) as \(h=\frac{v_{0}^{2}}{2g}\).
Substitute \(v_0 = 12\space m/s\) and \(g = 10\space m/s^{2}\) into the formula:

$$h=\frac{12^{2}}{2\times10}=\frac{144}{20}=7.2\space m\approx7.3\space m$$

Answer:

C. \(3.0\space s\)

Problem 14