Question

A body is thrown vertically up from ground with initial velocity 10m/s.Calculate the maximum height attained by the body if air friction force varies with velocity as |F| = mv^2​

Answer 1

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Answer 2

Question: A body is thrown vertically up from ground with initial velocity 10 m/s. Calculate the maximum height attained by the body of air friction force varies with velocity as |F| = mv²

Given: A body is thrown vertically up from ground with initial velocity 10 m/s.

To Find: Calculate the maximum height attained by the body of air friction force varies with velocity as |F| = mv²

Solution: If a body is moving vertically downwards or upwards, it experiences a downward acceleration due to the gravitational force of the earth. This is called acceleration due to gravity and is denoted by the symbol $g$. Strictly speaking $g$ is not a constant, but varies from place to place on the surface of the earth and also with height. However the variation of $g$ is so small that it can be neglected and $g$ can be considered as a constant unless very large heights are involved. Therefore, we can use the equations of motion for constant acceleration.

→ Important point: For solving problems of vertical motion under gravity, either the upward or the downward direction is taken as positive, then $g$ becomes negative and vice-versa. The signs of other quantities, $u, v$ and $x$ are decided according to the given problem.

If a body is thrown upwards with a velocity $u,$ then neglecting the effect of air resistance, the following results can be obtained using the equations of motion:

Maximum height attained: $h=\frac{u^{2} }{g}$

Time to reach the highest point: $t_1=\frac{u}{g}$

Total time taken to reach the top and then return to the ground: $t=\frac{2u}{g}$

Speed with which the body returns to the ground: $v=u$

Time taken to reach the ground: $t=\sqrt{\frac{2h}{g} }$

Velocity with which the body reaches the ground: $v=\sqrt{2gh}$