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Rashid [163]
3 years ago
11

Calculate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high). **** (use

the data from above)
a 0.00 s
b 3.94 s
c 8.80 s
d 846 s
Physics
1 answer:
Shtirlitz [24]3 years ago
3 0

Answer:

c 8.80 s

Explanation:

Given parameters:

Height of Empire State Building  = 380m

Unknown:

Time taken for King Kong to fall  = ?

Solution:

To solve this problem, we use one of the kinematics equation.

The appropriate one is given below;

      H  = ut + \frac{1}{2} g t²  

H is the height

t is the time taken

g is the acceleration due to gravity

            380  = (0 x t) + ( \frac{1}{2}  x 9.8 x t²)

           380  = 4.9t²  

             t²  = 77.55

              t = 8.8s

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Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

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m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

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\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

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