SATHEE: Kinematics Question 505

Kinematics Question 505

Question: A body is thrown horizontally from the top of a tower of height 5 m.It touches the ground at a distance of 10 m from the foot of the tower.The initial velocity of the body is $(g = 10 m s^{2})$ [EAMCET (Engg.) 2000]

Options:

A) $2.5 m s^{1}$

B) $5 m s^{1}$

C) $10 m s^{1}$

D) $20 m s^{1}$

Show Answer

Answer:

Correct Answer: C

Solution:

$S = u \times \sqrt{\frac{2 h}{g}}$

therefore $10 = u \sqrt{2 \times \frac{5}{10}}$

therefore $u = 10 m / s$

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Kinematics Question 505

Question: A body is thrown horizontally from the top of a tower of height 5 m.It touches the ground at a distance of 10 m from the foot of the tower.The initial velocity of the body is (g=10ms2) [EAMCET (Engg.) 2000]

Options:

Answer:

Solution:

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Question: A body is thrown horizontally from the top of a tower of height 5 m.It touches the ground at a distance of 10 m from the foot of the tower.The initial velocity of the body is $(g = 10 m s^{2})$ [EAMCET (Engg.) 2000]