Question

A 200 g ball is shot up a vertical

column of viscous liquid with an

initial velocity of 10 m/s Assuming

the force impeding the ball's motion

because of the viscosity of the liquid

is constant at 4.5 N, calculate how

high the ball goes before coming to

rest.

Answer 1

When a ball is hit upwards the maximum height of a projectile is given by

$H=u^2y/2g$

$u$_o $y$ =$u$_o $sin$$\theta$

mass=200$g$

=0.2kg

innitial velocity($u)$ =$10m/s$

in this case, we can get the normal force acting on a body vertically due to gravity.

$f$_m=$mg$

=$0.2*9.8$

=$1.96N$

viscosity of the liquid=$4.5N$

$tan\theta$ $=4.5/1.96$

$tan$ ^-1=2.2959

$\theta$ $=66.46$⁰

from $h$_max$=$ $u^2*sin(\theta$$)^2$ $/2g$

$=$ $10^2*sin(66.46)^2$ $/2*9.8$

$=4.2884m$