SATHEE: Kinematics Question 415

Kinematics Question 415

Question: From a pole of height 10 m, a stone is thrown vertically upwards with a speed 5 m/s. The time taken by the stone, to hit the ground, it’s n times that taken by it to reach the highest point of it’s path. The value of n is $[t a k e g = 10 m / s^{2}]$

Options:

A) 2

B) 3

C) 4

D) 5

Show Answer

Answer:

Correct Answer: C

Solution:

Given, $H = 10 m$ , $u = 5 m / s, g = 10 m / s^{2}$

Speed on reaching ground $v = \sqrt{u^{2} + 2 g h}$ Now, v = u + at Now, $v = \sqrt{u^{2} + 2 g h} = - u + g t$

Time taken to reach highest point is $t = \frac{u}{g}$ ,

$\Rightarrow t = \frac{u + \sqrt{u^{2} + 2 g H}}{g} = \frac{n u}{g} (from question)$

$\Rightarrow 2 g H = n (n - 2) u^{2}$

$\Rightarrow m (n - 1) = \frac{2 g H}{u^{2}} = \frac{2 \times 10 + 10}{5 \times 5} = 8 \Rightarrow n = 4$

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Kinematics Question 415

Question: From a pole of height 10 m, a stone is thrown vertically upwards with a speed 5 m/s. The time taken by the stone, to hit the ground, it’s n times that taken by it to reach the highest point of it’s path. The value of n is [takeg=10m/s2]

Options:

Answer:

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Question: From a pole of height 10 m, a stone is thrown vertically upwards with a speed 5 m/s. The time taken by the stone, to hit the ground, it’s n times that taken by it to reach the highest point of it’s path. The value of n is $[t a k e g = 10 m / s^{2}]$