Question

Question: In Fig. 13-18b, the scale on which thephysicist stand reads . How long will the cantaloupe take to reach the floor if the physicist drops it (from rest relative to himself) at a height ofabove the floor?

Short answer

Answer:

Time taken by cantaloupe to reach the floor ist =1.1s

Step-by-step solution

Listing the given quantities

Mass of physicist 60 kg

Scale reading 220 N

Initial velocity ${\text{v}}_{\text{0}}=0m/s$

Distance to reach is 2.1 m

Understanding the concept of acceleration 

Using the given data in the problem, we can find the gravity. Moreover, by using that acceleration, we can find its time to reach the floor from rest.

Formula:

$$\begin{gathered} \text{F}=\text{mg} \\ \text{s}={\text{v}}_{\text{0}}\text{t}+\frac{\text{1}}{\text{2}}{\text{gt}}^{\text{2}} \end{gathered}$$

Calculation of the time taken by cantaloupe to reach the floor

We can write from Newton’s second law,

$$\begin{gathered} F=mg \\ 220N=60 kg \\ g=3.66 \end{gathered}$$

By using the kinematic equation,

$$\begin{gathered} d\text{s}={\text{v}}_{\text{0}}\text{t}+\frac{\text{1}}{\text{2}}{\text{gt}}^{\text{2}}\backslash \mathrm{hfill} \\ s=0m/s\times t+\frac{\text{1}}{\text{2}}{\text{gt}}^{\text{2}} \\ t=\sqrt{\frac{2s}{g}} \\ t=1.1s \end{gathered}$$

Time taken by cantaloupe to reach the floor is t = 1.1s