a stone is dtropped from the top of 40m heigh calculate its speed after 2 seconds .Also find the speed ith the stone strikes the ground.
Actually it is wrong....
S(distance)=40m
Speed= distance/time
i.e 40/2=20m/s.
To calculate the speed of the stone after 2 seconds, we can use the formula of motion under constant acceleration:
v = u + at
Where:
v = final velocity (speed)
u = initial velocity (0 m/s because the stone starts from rest)
a = acceleration due to gravity (-9.8 m/s^2, taken as negative because it acts downwards)
t = time (2 seconds)
Let's calculate the speed of the stone after 2 seconds:
v = 0 + (-9.8) * 2
v = -19.6 m/s
Therefore, the speed of the stone after 2 seconds is 19.6 m/s downward (taking the negative sign into account).
To find the speed when the stone strikes the ground, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (speed)
u = initial velocity (0 m/s because the stone starts from rest)
a = acceleration due to gravity (-9.8 m/s^2, taken as negative because it acts downwards)
s = displacement (40 m downwards, since the stone falls from a height of 40 m)
Let's calculate the speed when the stone strikes the ground:
v^2 = 0^2 + 2 * (-9.8) * (-40)
v^2 = 0 + 784
v = square root of (784)
v ≈ 28 m/s
Therefore, the speed at which the stone strikes the ground is approximately 28 m/s downward.
Let V be the velocity (positive downwards) and Y be the distance the stone has fallen (also positive downwards).
V = g t
g is the acceleration of gravity, which is 9.8 m/s^2
When it strikes the ground,
Y = 40 = (1/2) g t^2
t = sqrt (2 Y/g)= 2.857 s
V = sqrt (2 g Y) = g * t = 28 m/s