A ball is dropped from the height of 10m, find its velocity when it reaches the ground
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The distance is 45m. (s)
Acceleration is 9.8 m/s^2 (acceleration due to gravity). (a)
We have to find the time (t)
By the 3rd equation of motion,
s = ut + 1/2 at^2 (here u is the initial velocity of the ball)
Since the initial velocity was 0 m/s, the formula becomes s = 1/2 at^2.
(substituting the values in the formula)
10 = 1/2 x 9.8 x t^2
10 = 9.8/2 x t^2
10 = 4.9 x t^2 (9.8/2 = 4.9)
(divide 4.9 on both sides)
10 /4.9 = 4.9 x t^2 / 4.9
2.04 = t^2
1.42 = t
So it takes approximately 3 seconds for the ball to reach the surface of the earth (neglecting air resistance).
Hope this will help you.... ✌
Acceleration is 9.8 m/s^2 (acceleration due to gravity). (a)
We have to find the time (t)
By the 3rd equation of motion,
s = ut + 1/2 at^2 (here u is the initial velocity of the ball)
Since the initial velocity was 0 m/s, the formula becomes s = 1/2 at^2.
(substituting the values in the formula)
10 = 1/2 x 9.8 x t^2
10 = 9.8/2 x t^2
10 = 4.9 x t^2 (9.8/2 = 4.9)
(divide 4.9 on both sides)
10 /4.9 = 4.9 x t^2 / 4.9
2.04 = t^2
1.42 = t
So it takes approximately 3 seconds for the ball to reach the surface of the earth (neglecting air resistance).
Hope this will help you.... ✌
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