Rectilinear Motion

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Question

A balloon rises from rest on the ground with constant acceleration 1.25 m s^-2. A stone is dropped when the balloon has risen to a height of 40 m. Find the time ( in sec) taken by the stone to reach the ground.

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Solution

$h = \frac{1}{2} {at}^{2} \Rightarrow t = \sqrt{\frac{2 h}{a}} = \sqrt{\frac{2 \times 40 \times 8}{10}} = 8 s$
Velocity after 8 s, $v = 0 + 1.25 \times 8 = 10 {ms}^{- 1}$
$40 = - 10 t + \frac{1}{2} 10 t^{2}$
$10 t^{2} - 2 \times 10 t - 2 \times 40 = 0$
$t^{2} - 2 t - 8 = 0$
$t - 4 t + 2 t - 8 = 0$
$t = 4 s$

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Rectilinear Motion

Remember concepts with our Masterclasses.

A balloon rises from rest on the ground with constant acceleration 1.25 m s-2. A stone is dropped when the balloon has risen to a height of 40 m. Find the time ( in sec) taken by the stone to reach the ground.

h=12at2⇒t=2ha=2×40×810=8sVelocity after 8 s, v=0+1.25×8=10ms−140=−10t+1210t210t2-2×10t-2×40=0t2−2t−8=0t−4t+2t−8=0t=4s

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A balloon rises from rest on the ground with constant acceleration 1.25 m s^-2. A stone is dropped when the balloon has risen to a height of 40 m. Find the time ( in sec) taken by the stone to reach the ground.