Stone is thrown vertically upwards with a velocity of 4.9m/s calculate the maximum height reached ? the time taken to reach the maximum height ?
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Using the equations of motion -
• v^2 = u^2 + 2aS
At the highest point, velocity is 0. And as the stone is going vertically upwards acceleration will be gravitation acceleration with negative sign because of opposite direction.
So, v = 0 and a = - g
Therefore,
= > 0^2 = u^2 + 2aS
= > - u^2 = 2aS
= > - (4.9)^2 = 2(-g)S
= > - 24.01 = - 2(9.8)S ... { g = 9.8 }
= > 24.01/2(9.8) = S
= > 1.225 = S
Hence the maximum height reached is 1.225 m.
Using v = u + at
= > 0 = 4.9 - gt
= > gt = 4.9
= > (9.8)t = 4.9
= > t = 4.9/9.8 = 0.5 sec
Time taken is 0.5 s.
• v^2 = u^2 + 2aS
At the highest point, velocity is 0. And as the stone is going vertically upwards acceleration will be gravitation acceleration with negative sign because of opposite direction.
So, v = 0 and a = - g
Therefore,
= > 0^2 = u^2 + 2aS
= > - u^2 = 2aS
= > - (4.9)^2 = 2(-g)S
= > - 24.01 = - 2(9.8)S ... { g = 9.8 }
= > 24.01/2(9.8) = S
= > 1.225 = S
Hence the maximum height reached is 1.225 m.
Using v = u + at
= > 0 = 4.9 - gt
= > gt = 4.9
= > (9.8)t = 4.9
= > t = 4.9/9.8 = 0.5 sec
Time taken is 0.5 s.
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