A stone is dropped from a 100m high tower.How long does its take to fall? a) the first 50m and b) The second 50m
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If we let that g = 10 m/s² then using the second equation of motion.
h = ut + (1/2) gt²
Here, initial velocity u = 0 m/s , h = 50 m (for first part of question) g = 10m/s² and t =?
= 50 = 0 t + (1/2) 10 t² = 50 = 0 + 5t²
= 50 = 5t² = t² = 50/5
= t² = 10 = t = √10 s
= t = 3.16 s
For second part we first calculate the time for the ball to travel whole 100m using same equation.
Thus we have 100 = 0 t + (1/2) 10 t²
= 100 = 0 + 5t² = 100 = 5t²
= t² = 100/5 = t² = 20
= t = √20 = t = 4.47 s
Thus time take to cover the next 50 m will be found by subtracting the time for first fifty seconds from
total time to cover whole distance.
Thus, time taken to cover second 50 m = 4.47 - 3.16 = 1.31 s
h = ut + (1/2) gt²
Here, initial velocity u = 0 m/s , h = 50 m (for first part of question) g = 10m/s² and t =?
= 50 = 0 t + (1/2) 10 t² = 50 = 0 + 5t²
= 50 = 5t² = t² = 50/5
= t² = 10 = t = √10 s
= t = 3.16 s
For second part we first calculate the time for the ball to travel whole 100m using same equation.
Thus we have 100 = 0 t + (1/2) 10 t²
= 100 = 0 + 5t² = 100 = 5t²
= t² = 100/5 = t² = 20
= t = √20 = t = 4.47 s
Thus time take to cover the next 50 m will be found by subtracting the time for first fifty seconds from
total time to cover whole distance.
Thus, time taken to cover second 50 m = 4.47 - 3.16 = 1.31 s
sruthimuthu:
wlcm
Answered by
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- solu :- let the g is 10 m/s
then given
height is 100 m
so easy to solve this
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