A body is dropped from rest from a height of 8m. How long does it take to reach the ground?
A body is dropped from rest at a height of 8m.how long does it take to reach the ground?(g=10m²
To find the time it takes for the body to reach the ground when dropped from rest, we can use the equation for motion under constant acceleration.
The equation that relates the distance fallen, the acceleration due to gravity, and the time taken is:
s = ut + (1/2)gt^2
Where:
s is the distance fallen (8m in this case),
u is the initial velocity (0 m/s for a body at rest),
g is the acceleration due to gravity (-9.8 m/s^2, assuming downward direction),
and t is the time taken.
Since the body is dropped from rest, the initial velocity is zero, so we can simplify the equation to:
s = (1/2)gt^2
Substituting the known values:
8 = (1/2)(-9.8)t^2
Rearranging the equation:
16 = -9.8t^2
Dividing both sides by -9.8:
t^2 = -16/9.8
Taking the square root of both sides:
t ≈ √(-16/9.8)
We can discard the negative root since time cannot be negative.
Therefore, the approximate time it takes for the body to reach the ground is t ≈ √(16/9.8) seconds.
d=1/2 g t^2
solve for time t.