Question

A stone is thrown vertically upward with an initial velocity
of 30 m/s. Find the maximum height reached by the stone.
Also calculate the total distance covered by the stone.
(Take g= 10 m/s2

Answer 1

Explanation:

At the highest point,

v = final velocity = 0

u = initial velovity = 30 m/s

- g = acceleration upward

Using equation of motion,

=> v² = u² + 2aS

=> 0² = 30² + 2(-g)S

=> 0 = 900 + 2(-10)S

=> 45 = S

Maximum height reached is 45 m.

It covers the same distance(when coming downwards).

Total distance = 45m + 45m = 90m

Answer 2

Answer

h=u^2 / 2g

Explanation:

take height as h before coming down

kinematical eqn of motion are

v= u plus at

s=ut plus 1/2 at^2

v-u=2as