Question

The distance travelled by a freely falling body in the first three seconds is equal to the distance travelled in the last second of its free fall. The time of its free fall is ​​

Answer 1

Answer:

same as the e distance travel in first three second and last two second

Answer 2

Given:

• Initial Velocity (u) = 0 m/s
• Acceleration = g = 10 m/s²
• Displacement in 3 seconds = Displacement in the last second

To Find:

• Total time of free fall by the object.

Solution:

Using the Second Equation of motion we can calculate the displacement in the first 3 seconds. Hence we get,

$\begin{gathered}: \implies \sf s = ut + 0.5 at² \\ : \implies \sf \: s = 0(3) + 0.5 × (10) × (3)² \\ : \implies \sf s = 0 + 5 × 9 \\ : \implies \boxed{\frak{ s = 45 m}} \blue \bigstar\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \Big\{ {eq}^{n} \: 1 \Big\}\end{gathered}$

$\underline {\sf{Hence, the \: displacement \: covered \: in \: the \: first \: 3 \: seconds \: is \: \textsf{ \textbf{45 m.}}}}$

Now,

• Let us assume the last second to be 't'.

We know that, the formula to calculate displacement traversed in a nth second is given as:

$\boxed{ \frak { \color{navy}{s_n = u + \dfrac{1}{2} \times a \times (2n-1)}}} \red\bigstar$

Substituting n=t and all the given information we get:

$: \implies \sf s(t) = 0 + 0.5 × 10 × (2t - 1) \\ :\implies \sf \: s(t) = 5 ( 2t - 1 ) \\ : \implies\boxed{ \frak{ s(t) - 10t - 5}} \blue \bigstar   \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf\Big\{{eq}^{n} \: 2 \Big\}$

According to the question, (i) is equal to (ii). Hence we get:

$\begin{gathered} : \implies \sf 45 = 10t - 5 \\: \implies \sf 45 + 5 = 10t \\ : \implies \sf50 = 10t \\ : \implies \sf \: t = \frac{50}{10} \\ : \implies \underline{\boxed{ \frak{ t = 5 \: seconds}}} \pink\bigstar\end{gathered}$

$\underline {\sf{Hence, the \: total \: time \: of \: fall \: is \: \textsf{\textbf{t}} \: \sf{which \: is \: \textsf{\textbf{5}} \: seconds.}}}$

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