Question

A STONE IS TROWN VERTICALLY UPWARD WITH AN INITIAL VELOCITY v0 THE DISTANCE TRAVELLED BY IT IN TIME 1.5 v0 /g

Answer 1

Answer:Initial velocity u.

Acceleration = -g.

Time 4u/g.

Final velocity is using v = u + at,

v = u - g * (4u/3g) = -u/3.

Average velocity = [ u + (-u/3)]/2 = u/3

Distance = average velocity * time =( u/3)*( 4u/3g)=4 u^2/ 9g

Or

Using s = ut + 0.5 at^2

s = u *(4u/3g) - 0.5 *g (16u^2/9g^2) = 4u^2/ 9g

The above answer gives the displacement.

To find the distance,we should consider speed instead of velocity.

Vertical height is h = u^2/ (2g) and time taken is u/g.

The remaining time is ( 4u / 3g) - u/g = u /(3g)

Distance traveled in time u /(3g) is 0.5 g t^2

= 0.5 g u^2/ ( 9g^2)= u^2/18g

Total distance is u^2/ (2g) + u^2/18g = 5u^2/(9g.)