Question

Questions - A solid ball is gently dropped from 80m height . It is found that velocity of ball increases uniformly at the rate of 10 ms2 With what velocity will it stike the ground? And After what time will it stirke the ground ?

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Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\bf\Large\boxed{40m/s,\:4sec}$

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$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

✭Initial velocity (u) = 0

✭Final velocity (v) = ?

✭Height of fall ,s= 80 m

✭Acceleration, a = 10 m/s²

✭Time of fall, t = ?

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As we know

${v}^{2} - {u}^{2} = 2as$

$= > {v}^{2} - 0 = 2 \times 10 \times 80$

=> v = 40 m/s

Therefore velocity while strike the ground is 40 m/s

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Now for t we know

v = u + at

⇒ 40 = 0 + 10 t

⇒t = 40/10

⇒t = 4 sec

Hence, time of fall is 4 sec

Answer 2

$\huge\boxed{\texttt{\fcolorbox{red}{aqua}{Answer :-}}}$

As Per Given

s=80 m

u= 0m/s (Because Initial Ball Is At Rest)

a= 10m/s2

$\mathfrak{\huge{\purple {\underline{\underline{To Find :}}}}}$

Final Velocity [v] And Time Taken[t]

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As We Known

Newton's 3rd Equation Of Motion

${v}^{2} - {u}^{2} = 2as \\ {v}^{2} - {0}^{2} = 2 \times 10 \times 80 \\ {v}^{2} = 1600 \\ v = \sqrt{1600} \\ v = 40 \: m {s}^{ - 1}$

And In Order To Find Time used

Newton's 1st Equation Of Motion

$v = u + at \\ 40 = 0 + (10)t \\ 40 = 10t \\ t = 4 \: sec$

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