Physics, asked by Anonymous, 12 hours ago

A body is freely falling. The displacement in the last second is equal to the displacement in the first three
seconds. What is the time of fall?​

Answers

Answered by ashishrockspj
0

Explanation:

Distance travelled in first 3 seconds=(1/2)gt

2

=(1/2)×9.8×9=44.1m

Let t be the time taken to travel the total height;

velocity at (t-1)th second =g×(t−1);

distance travelled at the last second =g(t−1)−(1/2)g=g×(t−0.5)=44.1;

hencet=

9.8

44.1

+0.5=5s

Answered by Anonymous
2

Answer:

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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