An object is dropped from the top of a building and is observed to take 7. 2s to hit the ground. How tall is the building?.
1 answer:
the height of the building is H=36 m.
<h3>What is The Law of Gravity?</h3>According to Newton's law of gravity, every particle of matter in the universe is attracted to every other particle with a force that varies directly as the product of their masses and inversely as their distance from one another.
Properties of Gravity -
- It is a universal attractive force. It is directly proportional to the product of the masses of the two bodies.
- It obey inverse square law.
- It is the weakest force known in nature.
Examples of Gravity -
- The force that holds the gases in the sun.
- The force that causes a ball you throw in the air to come down again.
- The force that causes a car to coast downhill even when you aren't stepping on the gas.
v₀=0 m/s
H₀=0 m
g=10 m/s²
t=7,2 s
H - ?
H = 0 +0 × 7.2 + 10(7.2)²/2
H = 36m
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Answer:
Diagrams in pictures
Explanation:
Using energy I can get
m g h = 1/2 m v^2
So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.
(H= 1,3m sin25)
V=2,32m/s
V= a t
And
X= v t +1/2 a t^2
Knowing v=2,32 m/s and x= 1,3 m
I can get
a= 6,21m/s2
F= m a
I can get the force of the box when it collides with the spring
F= 12, 425 N
The force the spring makes on the box then is
F = -12,425N = -k d
Then the spring's constant is k= 51,75N/m
To make the two diagrams I need the functions of time when the box slows down
I use the same two equations
V= a t
And
X= v t + 1/2 a t^2
Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.
I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).
Then,
V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec
X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.
for time between 0 and 0,689 sec
Diagrams and equations are in the pictures
