SATHEE: Kinematics Question 225

Kinematics Question 225

Question: A balloon starts r it’sing from the ground with an acceleration of 1.25 m/s2 after 8s, a stone is released from the balloon. The stone will ( $g = 10$ m/s2) [KCET 2001]

Options:

A) Reach the ground in 4 second

B) Begin to move down after being released

C) Have a displacement of 50 m

D) Cover a distance of 40 m in reaching the ground

Show Answer

Answer:

Correct Answer: A

Solution:

When the stone is released from the balloon. it’s height $h = \frac{1}{2} a t^{2} = \frac{1}{2} \times 1.25 \times {(8)}^{2} = 40 m$ and velocity $v = a t = 1.25 \times 8 = 10 m / s$ Time taken by the stone to reach the ground $t = \frac{v}{g} [1 + \sqrt{1 + \frac{2 g h}{v^{2}}}] = \frac{10}{10} [1 + \sqrt{1 + \frac{2 \times 10 \times 40}{{(10)}^{2}}}]$ =4 sec

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Kinematics Question 225

Question: A balloon starts r it’sing from the ground with an acceleration of 1.25 m/s2 after 8s, a stone is released from the balloon. The stone will ( g=10 m/s2) [KCET 2001]

Options:

Answer:

Solution:

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Question: A balloon starts r it’sing from the ground with an acceleration of 1.25 m/s2 after 8s, a stone is released from the balloon. The stone will ( $g = 10$ m/s2) [KCET 2001]