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

a)

b)

Explanation:

Given:

The mass of the stone:

Now the acceleration of the stone opposite of the motion:

where:

d = deceleration

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

a)

Now using the equation of motion:

where:

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

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

b)

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

initial velocity,

The acceleration will be:

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:

(+ve acceleration because it acts in the direction of motion)

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

double the distance = divide strength by 4