Question

A body is dropped from a height of 125m and takes some time to reach the ground (g = 10 m/s2) Find the time taken​

Answer 1

Okay, for this question we use kinematics equation : $\large D = V_{i} t + \frac {1}{2} at^{2}$

Then we solve for t, however, we can cancel out $\large V_{i} t$ because $\large V_{i}$ is going to be 0 (the moment you release the object it has no velocity) and anything times 0 is 0. So:

Multiply by 2:

$\large 2D = at^{2}$

Then get the square root of both sides so that you isolate t:

$\large \sqrt{ \frac{ 2D }{ a } } = t$

Plug in the numbers (keeping in mind a is -10 m/s/s and D is -125 m) :

$\large \sqrt{ \frac{ 2 \times -5 }{ -10 } }$

$\large \sqrt{25} = t$

t = 5

However, since time cannot be negative, we then rule out the negative answer and all we are left with is 5 s.

Answer 2

ANSWER:-

5sec.

EXPLANATION:-

A body is dropped from a height,

Initial velocity,u=0m/s

Distance,s=125m

Acceleration,a=10m/s^2

Time taken,t=?

Using 2nd equation(position-time relation),

S=ut+1/2at^2

Substitute the given values,

125m=0×t+1/2×10×t^2

125m=1/2×10m/s^2×t^2

125m=5m/s^2×t^2

t^2×5m/s^2=125m

t^2=125m/5ms^-2

t^2=25sec^2

t=√25s^2

t=5sec

Therefore ,time taken is 5 sec.

HOPE IT HELPS!!!...