Question

From the top of a Cliff 100m above the sea level a stone is thrown vertically upward with a velocity of 30m/s after how many seconds and with what velocity will it hit the sea

Answer 1

Given,

• Initial Velocity (u) = -30 m/s
• Acceleration (a) = 10 m/s²
• Distance (s) = 100 m

We know the equation:

$\boxed{\bold{\sf{v^2=u^2+2as }}}$

Applying the values to the equation:

$\sf{\implies v^2=(-30)^2+(2\times 10\times 100) }$

$\sf{\implies v^2=900+2000 }$

$\sf{\implies v^2=2900 }$

$\sf{\implies v=53.85 \ m/s }$

$\boxed{\bold{\sf{\therefore Velocity=53.85 \ m/s }}}$

We know the equation:

$\boxed{\bold{\sf{v=u+at }}}$

Applying the values to the equation:

$\sf{\implies 53.85=-30+10t }$

$\sf{\implies 10t = 53.85+30 }$

$\sf{\implies 10t = 83.85 }$

$\sf{\implies t = \dfrac{83.85}{10} }$

$\sf{\implies t = 8.385 \ s }$

$\boxed{\bold{\sf{\therefore Time \ Taken=8.385 \ s }}}$

$\huge{\boxed{\bold{\sf{\therefore Velocity=53.85 \ m/s }}}}$

$\huge{\boxed{\bold{\sf{\therefore Time \ Taken=8.385 \ s }}}}$

Answer 2

Given

• Initial height = 100 m
• Initial velocity = 30 m/s
• g = 10 m/s²

To Find

• Time taken and final velocity

Solution

☯ v²-u² = 2as & v = u + at

• Assuming downward motion to be positive

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✭ According to the Question :

⇒ v²-u² = 2as

⇒ v²-30² = 2(10)(100)

⇒ v² = 900+2000

⇒ v² = 2900

⇒ v = √2900

⇒ v = 10√29 m/s

━━━━━━━━━━━━━━━━━━━━

So then the time taken would be,

⇒ v = u + at

⇒ 10√29 = (-30) + 10t

⇒ (10√29+30)/10 = t

⇒ t = (√29+3) sec