Question

35. A ball is dropped from the top of a building 490 m
high. How long will it take to reach the ground ? What
will be its velocity when it strikes the ground ?​

Answer 1

Answer:

use newton's three equation.

Explanation:

t aayega 10 sec dusri equation se and velocity 1st equation se 98 m per sec

Answer 2

$\huge{\red{\bold{\underline{\underline{Answer:}}}}}$

$\green{\bold{\underline{\underline{Time=10 s}}}}$

$\green{\bold{\underline{\underline{Final\: Velocity=9.81 m/s}}}}$

$\huge{\red{\bold{\underline{\underline{Solution:}}}}}$

$Given \\ h \: = \: 490 \: m \\ u \: = \: 0 \: m {s}^{ - 1} \\ g \: = \: 9.81 \: m {s}^{ - 2}$

$Now \: by \: using \: height \: velocity \: equation:- \\ h \: = \: ut \: + \: \frac{1}{2} \: g {t}^{2}$

$where \: u \: = \: 0 \\ Therefore, \\ we \: use \: \\ h \: = \: \frac{1}{2} \: g {t}^{2}$

$Putting \: the \: given \: values \: into \: equation$

$= > \: 490 = \: \frac{1}{2} \times 9.81 \: {t}^{2}$

$= > \: {t}^{2} \: = \frac{490}{4.9} = 100$

$= > \: t \: = \sqrt{100} = 10$

$Hence, \: t \: = 10 \: s$

$Now \\ v \: = \: u + gt \\ = > 0 + 9.81 \times 10 = 9.81 \: m {s}^{ - 1}$