Question

A STONE IS THROWN VERTICALLY UPWARD WITH AN INITIAL VELOCITY V0 . THE DISTANCE IN TIME 4VO/3g IS

Answer 1

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.)

Answer 2

Answer:

The distance covered by stone in given time is

$S=\frac{4V_{0} {}^{2} }{9g}$

Explanation:

Given that,

A stone is thrown vertically upward with an initial velocity of V0 m/s.We are required to find the distance it covers in the time

$t=\frac{4V_{0}}{3g}$

For that,we use the kinematic equation of motion

$S=ut+\frac{1}{2}at^{2}$

Here,

Acceleration a=-g

Initial velocity u and time t are given.Hence,substituting them in the above equation,we get

$S=V_{0}( \frac{4V_{0}}{3g} ) - \frac{1}{2} g( \frac{4V_{0}}{3g} ) {}^{2} \\ = \frac{4V_{0} {}^{2} }{3g} - ( \frac{16 {V_{0}}^{2} }{9g {}^{2} } ) \times \frac{g}{2} \\ = \frac{4V_{0} {}^{2} }{3g} - \frac{8V_{0} {}^{2} }{9g} \\ = \frac{4V_{0} {}^{2} }{9g}$

Therefore,the distance covered by stone in the given time is

$S=\frac{4V_{0} {}^{2} }{9g}$

#SPJ3