Question

A ball is dropped from the edge of a building. If it hits the ground with a speed of 39 m/s, how much time had it been falling for, in units of seconds?

Answer 1

Answer:

${\bold{\therefore Time\:taken=3.9\:sec}}$

Explanation:

${\bold{\huge{\underline{SOLUTION-}}}}$

• In the given question information given about a ball that is dropped from a building and its final velocity is given.

• We have to find time taken to reach ground.

$\underline \bold{Given : } \\ \implies Initial \: velocity(u) = 0 \\ \implies Final \: velocity(v) = 39 \: m {s}^{ - 1} \\ \implies Acceleration(a) = g = 10 \: m {s}^{ - 2} \\ \\ \underline \bold{To \: Find : } \\ \implies Time = ?$

• According to given question :

$\bold{First \: method : } \\ \bold{By \: Third \: equation \: of \: motion : }\\ \implies {v}^{2} = {u}^{2} + 2as \\ \implies {39}^{2} = {0}^{2} + 2 \times 10 \times s \\ \implies s = \frac{1521}{20} \\ \bold{\implies s = 76.05 \: m} \\ \\ \bold{By \: Second \: equation \: of \: motion : } \\ \implies s = ut + \frac{1}{2} a {t}^{2} \\ \implies 76.05 = 0 \times t + \frac{1}{2} \times 10 \times {t}^{2} \\ \implies 76.05 = 5 {t}^{2} \\ \implies {t}^{2} = \frac{76.05}{5} \\ \implies t = \sqrt{15.21} \\ \bold{\implies t = 3.9 \: sec}$

$\bold {Alternate \: method : } \\ \bold{By \: First \: equation \: of \: motion : }\\ \implies v = u + at \\ \implies 39 = 0 + 10 \times t \\ \implies t = \frac{39}{10} \\ \bold{\implies t = 3.9 \: sec } \\ \\ \bold{\therefore Time \: taken \: to \: reach \: ground \: is \: 3.9 \: sec}$

Answer 2

$\huge{\mathfrak{\underline{\underline{\red{Answer :-}}}}}$

Time taken = 3.9 s

_________________________

$\mathrm{\underline{\gray{Given :-}}}$

Initial velocity(u) = 0 m/s

Final velocity(v) = 39 m/s

Acceleration due to gravity = 10 m/s²

__________________________

$\mathrm{\underline{\gray{To Find :-}}}$

Time taken (t)

___________________________

$\mathrm{\underline{\gray{Solution :-}}}$

By third equation of motion

$\huge{\boxed{\boxed{v^{2} \: = \: u^{2} + 2gs}}}$

____________________[put values]

⇒ (39)² = (0)² + 2 * 10 * s

⇒ 1521 = 20 * s

⇒ 1524/20 = s

⇒76.05 m = s

$\huge{\boxed{\boxed{Distance(s) \: = \: 76.05 m}}}$

By second equation of motion

$\huge{\boxed{\boxed{s \: = ut + \frac{1}{2} at^{2}}}}$

______________[Put values]

⇒76.05 = 0 * t + 5 * t²

⇒ t² = 15.21

⇒ t = √15.21

⇒t = 3.9 s

$\huge{\boxed{\boxed{Time(t) \: = \: 3.9 s}}}$