Question

An object thrown vertically upwards reaches a height of 500 m. what was it's initial velocity? How long will the object take to come back to earth?​

Answer 1

initial velocity (u) = 100 m/s

time taken to come back (t) = 10+10 = 20 sec

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Answer 2

Given:-

• Height = h = 500m

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Need to find:-

• The initial velocity = u = ?
• Time taken by the object to reach Earth= Time of flight = T = ?

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Solution:-

We know, at maximum height,

• Final velocity = v = 0m/s
• For an object thrown vertically upwards, a = -g = -10 m/s²

From 3rd Equation of motion we can say,

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

$\implies {0}^{2} - {u}^{2} = 2 \times ( - 10) \times 500$

$\implies - {u}^{2} = - 10000$

$\implies {u}^{2} = 10000$

$\red {\bold{\boxed{\implies u = 100m/s}}}$

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We know, from 1st equation of motion

$s = ut + \dfrac{1}{2} a {t}^{2}$

$\implies 1000 = 100 \times t - 10 \times \dfrac{1}{2} {t}^{2}$

$\implies 1000 = 100t - 5 {t}^{2}$

$\implies 5 {t}^{2} - 100t + 1000 = 0$

$\implies {t}^{2} - 20t + 200 = 0$

$\implies v = u + at$

$\implies 0 = 100 + ( - 10)t$

$\implies 10t = 100sec$

$\red{\bold{\boxed{\implies t = 10sec}}}$