Question

you throw a glob of putty straight up toward the ceiling, which is 3.20 m above the point where the putty leaves your hand. the initial speed of the putty as it leaves your hand is 9.70 m/s. part a what is the speed of the putty just before it strikes the ceiling? express your answer with the appropriate units. part b how much time from when it leaves your hand does it take the putty to reach the ceiling? express your answer with the appropriate units.

Explanation:

Step1: Identify the kinematic - equation

The kinematic equation \(v_f^2=v_i^2 - 2gh\) is used for vertical - motion under gravity, where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(g = 9.8\ m/s^2\) is the acceleration due to gravity, and \(h\) is the height.

Step2: Calculate the speed before hitting the ceiling (Part A)

Given \(v_i=9.70\ m/s\), \(g = 9.8\ m/s^2\), and \(h = 3.20\ m\).

$$v_f=\sqrt{v_i^2 - 2gh}=\sqrt{(9.70)^2-2\times9.8\times3.20}$$

$$v_f=\sqrt{94.09 - 62.72}=\sqrt{31.37}\approx5.60\ m/s$$

Step3: Identify the kinematic - equation for time

The kinematic equation \(v_f=v_i - gt\) is used to find the time \(t\). We can re - arrange it to \(t=\frac{v_i - v_f}{g}\).

Step4: Calculate the time (Part B)

We know \(v_i = 9.70\ m/s\), \(v_f\approx5.60\ m/s\), and \(g = 9.8\ m/s^2\).

$$t=\frac{9.70 - 5.60}{9.8}=\frac{4.1}{9.8}\approx0.42\ s$$

Answer:

Part A: \(v_f = 5.60\ m/s\)
Part B: \(t = 0.42\ s\)