Question

A stone with a weight of 5.30 N is launched vertically from ground level with an initial speed of 23.0 m/s, and the air drag on it is 0.266 N throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground

Answer 1

Answer:

a) $h=25.7\ m$

b) $v'=21.8733\ m.s^{-1}$

Explanation:

Given:

• weight of the stone, $w=5.3\ N$

• initial velocity of vertical projection, $u=23\ m.s^{-1}$

• air drag acting opposite to the motion of the stone, $D=0.266\ N$

The mass of the stone:

$m=\frac{w}{g}$

$m=\frac{5.3}{9.8}$

$m=0.5408\ kg$

Now the acceleration of the stone opposite of the motion:

$D=m.d$

where:

d = deceleration

$0.266=0.5408\times d$

$d=0.4918\ m.s^{-2}$

In course of going up the net acceleration on the stone will be:

$g'=g+d$

$g'=9.8+0.4918$

$g'=10.2918\ m.s^{-2}$

a)

Now using the equation of motion:

$v^2=u^2-2 g'.h$

where:

$v=$ final velocity when the stone reaches at the top of the projectile = 0

h = height attained by the stone before starting to fall down

$0^2=23^2-2\times 10.2918\times h$

$h=25.7\ m$

b)

during the course of descend from the top height of the projectile:

initial velocity, $v=0\ m.s^{-1}$

The acceleration will be:

$g"=g-d$

$g"=9.8-0.4918$

$g"=9.3082\ m.s^{-2}$

here the gravity still acts downwards but the drag acceleration acts in the direction opposite to the motion of the stone, now the stone is falling down hence the drag  acts upwards.

Using equation of  motion:

$v'^2=v^2+2g".h$ (+ve acceleration because it acts in the direction of motion)

$v'^2=0^2+2\times 9.3082\times 25.7$

$v'=21.8733\ m.s^{-1}$