a stone is thrown vertically upward direction with a velocity of 5 metre per second 8 acceleration of the stone during its motion 10 metre per second square what will be the height attained by the stone and how much time will it reach there
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Answered by
115
Given :
Initial velcoity=u=5m/s
acceleration=10m/s2
time=t=?
Maximum height=h=?
formula to be used:
Maximum Height =Hmax=u²/2g
=5x5/2x10
=5/4m
=1.25m
Time of ascent= Time taken to reach maximum height=u/g
=5/10
=1/2
=0.5 sec
Initial velcoity=u=5m/s
acceleration=10m/s2
time=t=?
Maximum height=h=?
formula to be used:
Maximum Height =Hmax=u²/2g
=5x5/2x10
=5/4m
=1.25m
Time of ascent= Time taken to reach maximum height=u/g
=5/10
=1/2
=0.5 sec
Answered by
39
To find the height attained,
v2 = u2 + 2as
s = Height attained = (v2 - u2)/2a = (0 - 25)/(2 x -10) = 1.25 .
a is negative since it's direction is opposite to the direction of motion.
t = (v - u)/a = -5/-10 = 0.5 s.
v2 = u2 + 2as
s = Height attained = (v2 - u2)/2a = (0 - 25)/(2 x -10) = 1.25 .
a is negative since it's direction is opposite to the direction of motion.
t = (v - u)/a = -5/-10 = 0.5 s.
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