two objects of masses m1 and m2 fall freely from height h1 and h2 in a vertical plane respectively.The ratio of time taken by them is how much by reaching the ground?
Didn't Galileo do this experiment in Pisa?
h=1/2 g t^2
so t1/t2=sqrt(h1/h2)
m1/m2
To find the ratio of the time taken by two objects of masses m1 and m2 falling freely from heights h1 and h2 in a vertical plane respectively, we can use the fact that the time taken to fall freely from a certain height is independent of the mass of the object.
The time taken to fall freely can be calculated using the equation:
t = √(2h/g)
where,
t = time taken to fall,
h = height from which the object falls, and
g = acceleration due to gravity.
Therefore, the ratio of the time taken by the two objects can be calculated as:
Ratio = (t1 / t2)
= √(2h1 / g) / √(2h2 / g)
= √(h1 / h2)
Hence, the ratio of the time taken by the two objects is √(h1 / h2).
To find the ratio of time taken by two objects of masses m1 and m2 falling freely from heights h1 and h2 in a vertical plane respectively, we can use the concept of free fall and the equations of motion.
Let's assume that the object with mass m1 falls from height h1 and takes time t1 to reach the ground. Similarly, the object with mass m2 falls from height h2 and takes time t2 to reach the ground.
In free fall, neglecting air resistance, both objects will experience the same acceleration due to gravity (g), which is approximately 9.8 m/s^2. The distance fallen by an object can be calculated using the equation:
s = ut + (1/2)gt^2
where s is the distance fallen, u is the initial velocity (which is zero in the case of free fall), g is the acceleration due to gravity, and t is the time taken.
For the object with mass m1 falling from height h1, the distance fallen (s1) can be written as:
s1 = h1 = (1/2)g(t1^2) ----(1)
Similarly, for the object with mass m2 falling from height h2, the distance fallen (s2) can be written as:
s2 = h2 = (1/2)g(t2^2) ----(2)
We can rearrange equations (1) and (2) to isolate t1^2 and t2^2 respectively:
t1^2 = (2h1)/g ----(3)
t2^2 = (2h2)/g ----(4)
Now, we can find the ratio of time taken by the two objects by dividing equation (3) by equation (4):
(t1^2/t2^2) = [(2h1)/g] / [(2h2)/g]
(t1^2/t2^2) = (h1/h2)
Taking the square root of both sides, we get:
(t1/t2) = sqrt(h1/h2)
So, the ratio of time taken by the two objects to reach the ground is sqrt(h1/h2).