A car falls off a ledge and drops to the ground in $ 0.5 \mathrm{~s} $. Let $ g=10 \mathrm{~m} \mathrm{~s}^{-2} $ (for simplifying the calculations).

1. What is its speed on striking the ground?
2. What is its average speed during the $ 0.5 \mathrm{~s} ? $
3. How high is the ledge from the ground?

Given:

Time, $t$ = 0.5 sec

Acceleration, $a$ = +10m/s² (acceleration is positive as car falls downwards)

Initial velocity, $u$ = 0 m/s

(i) We need to find the final velocity, $v$.

We have $u$, $a$, and $t$.

Then we can find $v$, using first law of motion.

$v=u+at$

$v=0+10\times 0.5$

$v=5m/s$

Therefore, the speed of car on striking the ground is 5m/s.

(ii) We know that,

$Average\ speed=\frac{initial\ velocity+final\ velocity}{2}$

$Average\ speed=\frac{u+v}{2}$

$Average\ speed=\frac{0+5}{2}$

$Average\ speed=2.5m/s$

Hence, the average speed of car is 2.5 m/s.

(iii) Height of the ledge would be equal to the distance travelled, $s$.

We have $u$, $a$, and $t$.

Then we can find $s$, using the second law of motion.

$s=ut+\frac{1}{2}a{t}^{2}$

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

$s=0+\frac{1}{2}\times 10\times 0.25$

$s=5\times 0.25$

$s=5\times \frac{25}{100}$

$s=5\times \frac{1}{4}$

$s=1.25m$

Hence, the height of the ledge from the ground is 1.25m.

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Updated on: 10-Oct-2022

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