Answer :
Explanation:
Let us assume that the initial velocity is u. And, according to the kinematics,
s = [tex]ut + \frac{1}{2}gt^{2}[/tex]
ut = s - [tex]\frac{1}{2}gt^{2}[/tex]
Putting the given values into the above formula as follows.
ut = s - [tex]\frac{1}{2}gt^{2}[/tex]
[tex]u \times 0.15 s = 1.6 - \frac{1}{2}9.8 \times (0.15 s)^{2}[/tex]
u = 9.93 m/s
Hence, the required height will be calculated as follows.
h = [tex]\frac{u^{2}}{2g}[/tex]
= [tex]\frac{(9.93)^{2}}{2 \times 9.8}[/tex]
= 5.03 m
Thus, we can conclude that security look for the raucous hotel guests at 5.03 m.
Answer:
Security should look for the raucous hotel guests at height of [tex]5.03[/tex] meters
Explanation:
Given
Height of the window [tex]= 1.6[/tex] meters
Time taken by balloon to pass the window [tex]= 0.15[/tex] seconds
As per second law of motion for objects moving upward
[tex]h = u*t - \frac{1}{2} gt^2\\[/tex]
Substituting the given values we get -
[tex]1.6 = u * 0.15 - 0.5 * 9.8 * 0.15^2[/tex]
u = [tex]\frac{1.6 + 0.5 * 9.8 * 0.15^2}{0.15}[/tex]
[tex]u = 9.93 \frac{m}{s}[/tex]
Security should look for the raucous hotel guests at height [tex]"h"[/tex]
Here
[tex]h = \frac{v^2}{2* g}[/tex]
Substituting the values in above equation, we get -
[tex]h = \frac{9.93^2}{2 * 9.81} \\h = 5.03[/tex]
Security should look for the raucous hotel guests at height of [tex]5.03[/tex] meters