SATHEE: Kinematics Question 687

Kinematics Question 687

Question: The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8 t - 5 t^{2})$ meter and $x = 6 t$ meter, where $t$ it’s in second. The velocity with which the projectile is projected it’s

Options:

A) 8 m/sec

B) 6 m/sec

C) 10 m/sec

D) Not obtainable from the data

Show Answer

Answer:

Correct Answer: C

Solution:

[c] $v_{y} = \frac{d y}{d t} = 8 - 10 t$ , $v_{x} = \frac{d x}{d t} = 6$

at the time of projection i.e. $v_{y} = \frac{d y}{d t} = 8$ and $v_{x} = 6$

$∴ v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 m / s$

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Kinematics Question 687

Question: The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y=(8t−5t2) meter and x=6t meter, where t it’s in second. The velocity with which the projectile is projected it’s

Options:

Answer:

Solution:

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Question: The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8 t - 5 t^{2})$ meter and $x = 6 t$ meter, where $t$ it’s in second. The velocity with which the projectile is projected it’s