A 1.5 kg orange falls from a tree and hits the ground in 0.75s. What is the speed of the orange just before it hits the ground?
1 answer:
The final speed of the orange is 7.35 m/s
Explanation:
The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration towards the ground. So we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time elapsed
For the orange in this problem, we have
u = 0 (it is dropped from rest)
is the acceleration
Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:
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Answer:
1) The greatest height attained by the ball equals 20.387 meters.
2) The time it takes for the ball to reach 15 meters approximately equals 1 second.
Explanation:
The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.
thus using third equation of kinematics we obtain the height attained as
where
'v' is the final speed of the ball
'u' is the initial speed of the ball
'a' is the acceleration that the ball is under which in this case equals 9.81
's' is the distance it covers
Thus for maximum height applying the values in the equation we get
Using the same equation we can find the speed of the ball when it reaches 15 meters of height as
the time it takes to reduce the velocity to this value can be found by first equation of kinematics as
Answer:
The ratio is 9.95
Solution:
As per the question:
Amplitude,
Wavelength,
Now,
To calculate the ratio of the maximum particle speed to the speed of the wave:
For the maximum speed of the particle:
where
= angular speed of the particle
Thus
Now,
The wave speed is given by:
Now,
The ratio is given by: