Question

A tourist drops a rock from rest from a ground rail overlooking a valley.What is the velocity of the rock after 4.0sec?What is the displacement of the rock after 4.0 sec​

Answer 1

Explanation:

here acceleration is constant so we apply equation of motion

v=u+at

u=o

so v=at = 10×4=40 m/s

g= a=10

and displacement is

S=ut+1/2at^2

u=0

S= 1/2×10(4)^2 =800 m

Answer 2

ㅤ$\;\;\underline{\textbf{\textsf{Given:-}}}$

• Time taken, t = 4 sec

• Initial velocity, u = 0 m/s

• Acceleration, g = 10 m/s²

ㅤ$\;\;\underline{\textbf{\textsf{To Find :-}}}$

• Velocity

• Displacement

ㅤ$\;\;\underline{\textbf{\textsf{Solution :-}}}$

$\underline{\:\textsf{ Using 1st equation of motion :}}$

$\sf \dashrightarrow v = u + at$

$\dashrightarrow \sf v= 0+10\times 4\\ \\ \dashrightarrow \sf v= 40 m/s$

ㅤ$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ Velocity ( after 4 sec) </p><p>\textbf{ 40m/s}}}.$

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$\underline{\:\textsf{ Using 2nd equation of motion :}}$

$\sf \dashrightarrow s = ut+\dfrac{1}{2}at^{2}$

$\dashrightarrow \sf s= ut+\dfrac{1}{2}gt^2\\ \\ \dashrightarrow \sf s= 0\times 10 +\dfrac{1}{\cancel2}\times \cancel{10}\times (4)^2\\ \\ \dashrightarrow \sf s= 0+5\times 16\\ \\ \dashrightarrow \sf s= 80m$

ㅤ$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ The distance travelled is </p><p>\textbf{ 80m}}}.$

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