hi friends please solve the 13 the question
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Initial velocity (u) = 49 m/s Final velocity (v) = 0 m/s
Gravity in toward down = + 9.8 m/s2 Gravity in toward up = - 9.8 m/s2
(i) We know, V2 - u2 = 2gs
or, (o)2- (49)2 = 2 X 9.8 X S
or, s = - (49)2/ 2 X 9.8 = 122.5 m
Maximum height = 122.5 m
(ii) We know, v = u + gt
or 0 = 49 +(- 9.8) X t
or 9.8 X t = 49
or, t = 49/ 9.8 = 5 s
If, time for upward direction = time for downward direction
Then, total time taken by a ball to return back = 5 + 5 = 10 s
Gravity in toward down = + 9.8 m/s2 Gravity in toward up = - 9.8 m/s2
(i) We know, V2 - u2 = 2gs
or, (o)2- (49)2 = 2 X 9.8 X S
or, s = - (49)2/ 2 X 9.8 = 122.5 m
Maximum height = 122.5 m
(ii) We know, v = u + gt
or 0 = 49 +(- 9.8) X t
or 9.8 X t = 49
or, t = 49/ 9.8 = 5 s
If, time for upward direction = time for downward direction
Then, total time taken by a ball to return back = 5 + 5 = 10 s
Answered by
1
Hey there !
( i ) Initial Velocity = 49 m/s
Final Velocity = 0 m/s
Acceleration = - 9.8 m/s ( -g )
Formula = s = ut + at² / 2 => Formula 1
But we dont know time. So time can be calculated by using the formula,
v = u + at
=> 0 = 49 - 9.8 t
=> 9.8 t = 49
=> t = 49 / 9.8 = 5 seconds
So substituting all the values in Formula 1 we get,
s = 49 ( 5 ) - 9.8 * 5² / 2
=> s = 245 - 9.8 * 25 / 2
=> s = 245 - 245 / 2
=> s = 245 - 122.5
=> s = 122.5 m
Hence the ball would travel 122.5 m
( ii ) While going it takes 5 seconds and while coming it again takes 5 seconds. Hence Total time taken is 10 seconds.
Hope my answer helped !
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