A ball is thrown from a height h and another from 2h.find the ratio of time taken by the two balls to reach the ground.
H:2h
H:2h.. Its just a suggestion to get your answer
Smaj ni aya
To find the ratio of time taken by the two balls, we need to consider their motion and use the equations of motion.
Let's assume the initial velocity of both balls is the same, and neglect any air resistance. The free fall motion of a ball can be described by the equations:
h = (1/2)gt^2
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken.
For the first ball thrown from height h, the time taken to reach the ground is given by t1:
h = (1/2)gt1^2
For the second ball thrown from height 2h, the time taken to reach the ground is given by t2:
2h = (1/2)gt2^2
We can rearrange these equations to solve for t1 and t2:
t1 = sqrt(2h/g)
t2 = sqrt(4h/g) = 2sqrt(h/g)
Now, let's find the ratio of t1 to t2:
t1/t2 = sqrt(2h/g) / 2sqrt(h/g)
= sqrt(2h/g) * 1/(2sqrt(h/g))
= 1/(2^2 * sqrt(2))
= 1/(4 * sqrt(2))
= 1/(4√2)
Therefore, the ratio of time taken by the first ball to the second ball to reach the ground is 1/(4√2).