A car falls off a ledge and drops to the ground in 0.5 s. Let g = 10 ms-2. What is its speed on striking the ground and its average speed, how high is the ledge from the ground?

Given that, Acceleration, a = g = 10 m

s^{- 2}

Time of fall, t = 0.5 s
Let, initial velocity, u = 0
Now, Speed on striking the ground, v =

u + at = 0 + 10 \times 0.5 = 5.0 m s^{- 1}

Then, Average speed,

v_{av}

\frac{u + v}{2}

\frac{0 + 5.0}{2}

= 2.5 m

s^{- 1}

Hence, Height of the ledge, h = u t

+ \frac{1}{2} {at}^{2}

=
⸫ h =

0 \times 0.5 + \frac{1}{2} {\times 10 \times (0.5)}^{2}

= 1.25 m
So, the speed on striking the ground is 5.0 m

s^{- 1}

, average speed is 2.5 m

s^{- 1}

and the height of the ledge from the ground is

1.25 m

A car falls off a ledge and drops to the ground in 0.5 s. Let g = 10 m $s^{- 2}$ . What is its speed on striking the ground and its average speed, how high is the ledge from the ground?