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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).- What is its speed on striking the ground?
- What is its average speed during the $ 0.5 \mathrm{~s} ? $
- 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.