Question

a car falls off a ledge and drops to the ground in 0.5 seconds. Let g=10m/second. The height of the ledge from the ground is ​

Answer 1

Given:

Initial velocity of car, u= 0 m/s

Acceleration of car, g= 10 m/s²

Time taken by car, t= 0.5 s

To Find:

We know that,

• According to second equation of motion for constant acceleration,

$\orange{\boxed{\bf{s=ut+\dfrac{1}{2}at^{2}}}}$

where,

u is initial velocity

a is acceleration

s is displacement

t is time taken

$\rule{190}{1}$

Reference taken here:

All displacements, velocities, forces and accelerations acting in upward direction are taken positive.

All displacements, velocities, forces and accelerations acting in downward direction are taken negative.

$\rule{190}{1}$

Let the height of ledge be h

On applying second equation of motion on car, we get

$\longrightarrow\rm{s=ut+\dfrac{1}{2}at^{2}}$

$\longrightarrow\rm{-h=0(0.5)+\dfrac{1}{2}g(0.5)^{2}}$

$\longrightarrow\rm{-h=\dfrac{1}{2}(-10)(0.25)}$

$\longrightarrow\rm{-h=\dfrac{-2.5}{2}}$

$\longrightarrow\rm{-h=-1.25}$

$\longrightarrow\rm\green{h=1.25\:m}$

Hence, the height of the ledge is 1.25 m.

Answer 2

GiVeN :-

• Initial velocity (u) = 0 m/s

• Time taken (t) = 0.5 s

• Acceleration due to gravity (g) = 10 m/s²

To FiNd :-

The height(h) of the ledge from the ground.

AcKnOwLeDgEmEnT :-

We know the second law of Kinematics :-

$\boxed{\sf{h=ut+\frac{1}{2}gt^2 }}$

SoLuTiOn :-

Substituting the values in the above equation :-

$\displaystyle{\sf{h=0(0.5)+\frac{1}{2}\times 10 \times (0.5)^2 }}\\\\\displaystyle{\sf{h=0+5 \times 0.25 }}\\\\\boxed{\displaystyle{\sf{h=1.25\ m }}}$

Therefore, the height of the ledge is 1.25 m.

$\rule{198}2$

Sign convention :-

For our suitability, the acceleration has been taken positive. (i.e. direction to the downward position is positive, and direction to the upward position is negative).