Question

An object dropped from a cliff with a constant acceleration of 10 m/s . Find its speed 5 sec after it was dropped.

Answer 1

Answer:

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

$s = 0 \times 5 + \frac{1}{2} \times 10 \times {5}^{2}$

$s = 125 \: m$

$we \: know \: that \: \: speed = \frac{distance}{time}$

$speed = \frac{125}{5}$

Speed=25 m/s

Answer 2

Answer :

➥ The speed of an object after 5 sec = 50 m/s

Given :

➤ Intial velocity of an object (u) = 0 m/s

➤ Acceleration of an object (a) = 10 m/s²

➤ Time taken (t) = 2 sec

To Find :

➤ Final velocity of an object (v) = ?

Solution :

Speed 5 sec after it was dropped

From First equation of motion

$\tt{: \implies v = u + at}$

$\tt{:\implies v = 0 + 10 × 5}$

$\tt{: \implies \green{ \underline{ \overline{ \boxed{ \purple{ \bf{ \: \: v = 50 m/s \: \: }}}}}}}$

Hence, the speed of an object after 5 sec is 50 m/s.

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Some releted equations :

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as

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