Two balls are dropped from heights h and 2h respectively. What would b...
Concept:
For a body moving in a straight line with constant acceleration, then
- v = u + at --- (1)
- s = ut + 0.5at2 --- (2)
- v2 = u2 + 2as --- (3)
- Where u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time
Calculation:
Given that, for first ball, height, h1 = h and let time = t1
And for the second ball, height, h2 = 2h and time = t2
Since the object is falling freely then acceleration = g
Initial velocity for both the bodies, u = 0
Then from equation 2, for first ball
h1 = ut + 0.5gt12
⇒ h = 0 + 0.5gt12 --- (4)
For second ball,
h2 = ut + 0.5gt22
⇒ 2h = 0 + 0.5gt22 --- (5)
Dividing equation 4 by 5
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Two balls are dropped from heights h and 2h respectively. What would b...
Ratio of times taken by the balls to reach the earth:
- Let's assume the height from which the balls are dropped is h for the first ball and 2h for the second ball.
- The time taken by an object to fall from a certain height can be calculated using the formula: time = √(2h/g), where g is the acceleration due to gravity (approximately 9.8 m/s²).
Time taken by the first ball (h):
- For the first ball dropped from height h, the time taken to reach the earth can be calculated as: t1 = √(2h/g).
Time taken by the second ball (2h):
- For the second ball dropped from height 2h, the time taken to reach the earth can be calculated as: t2 = √(2(2h)/g) = √(4h/g) = 2√(h/g).
Ratio of times taken by the two balls:
- To find the ratio of times taken by the two balls, we divide the time taken by the second ball by the time taken by the first ball: t2/t1 = 2√(h/g) / √(2h/g) = 2√(h/g) / √(2h/g) = 2√(h/g) * √(g/2h) = 2√(h/2h) = 2√(1/2) = √2.
Therefore, the ratio of times taken by the two balls to reach the earth is √2, which corresponds to option b) 1 √2.