Question

A ball is thrown up with initial velocity $u$. The retardation due to the drag is proportional to the velocity with proportionality constant, $\beta$. The maximum height and the time taken to reach the maximum height are

• A. $\frac{u}{\beta }-\frac{g}{{\beta }^2}\ln (1+\frac{\beta u}{g}),\frac{1}{\beta }\ln (1+\frac{\beta u}{g})$
• B. $\frac{u}{\beta }-\frac{g}{{\beta }^2}\ln (1-\frac{\beta u}{g}),\frac{1}{\beta }\ln (1-\frac{\beta u}{g})$
• C. $\frac{u}{\beta }+\frac{g}{{\beta }^2}\ln (1-\frac{\beta u}{g}),\frac{1}{\beta }\ln (1+\frac{\beta u}{g})$
• D. $\frac{u}{\beta }-\frac{g}{{\beta }^2}\ln (1+\frac{\beta u}{g}),\frac{1}{\beta }\ln (1-\frac{\beta u}{g})$

Answer 1

The correct option is A $\frac{u}{\beta }-\frac{g}{{\beta }^2}\ln (1+\frac{\beta u}{g}),\frac{1}{\beta }\ln (1+\frac{\beta u}{g})$

Let the origin be at the ground and vertically up direction be positive. The acceleration of the ball is given by
$\frac{dv}{dt}=-g-\beta v=-g(1+\gamma v),......\left(1\right)$
where $\gamma =\frac{\beta }{g}$. Let the ball takes time $T$ to reach the maximum height $H$.
Initial velocity of the ball is $u$ and its velocity at the maximum height is zero. Integrate equation (1)
${\int }_u^0\frac{dv}{1+\gamma v}=-g{\int }_0^{T}dt$
to get $T=\frac{1}{\beta }\ln (1+\frac{\beta }{g}u).$
Re-write $\frac{dv}{dt}=v\frac{dv}{ds}$ in equation (1) and integrate
${\int }_u^0\frac{vdv}{1+\gamma v}=-g{\int }_0^{H}ds$
To get $H=\frac{u}{\beta }-\frac{g}{{\beta }^2}\ln (1+\frac{\beta }{g}u)$