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LoginA body is thrown at an angle `theta_0` with the horizontal such that it attains a speed equal to `(sqrt((2)/(3))` times the speed of projection when the body is at half of its maximum height. Find the angle `theta_0`.
General Awareness KINEMATICS IN GENERAL AWARENESS . 2 years ago
+10 pts
`v = sqrt(v_0^2 - 2 g y))`,
where `y = (y_(max))/(2)`
Sunstituting `y_(max) = (v_0^2 sin^2 theta_0)/(2 g)`, we have
`v = v_0 (sqrt(1 - (sin^2 theta_0)/(2)))`
Since `v = sqrt((2)/(3)) v_0` (given), we have `1 -(sin^2 theta_0)/(2) = (2)/(3)`
This given `sin theta_0 = sqrt((2)/(3))`. Hence, `theta_0 = sin^-1 (sqrt((2)/(3)))`.
Posted on 06 Oct 2022, this text provides information on General Awareness related to KINEMATICS IN GENERAL AWARENESS. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.
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