Question

A body is projected in vertically upward direction with a speed u .find the distance travelled by the body in reaching its top most point

Answer 1

Answer:

let h be the distance travelled by the body to reach top most point

h= ut - 1/2 a t^2

time required to reach top most point is u^2/2g

here g refers to gravitational force

a = g because there is only one acceleration and it is gravitational force

h = u(u/g) - 1/2g(u^2/g^2)

h = u^2/g -u^2/2g

h = u^2/2g

Explanation:

Answer 2

Answer:

Explanation:

Let the distance travelled by the body in reaching its upmost point be = s

therefore distance = s m

       initial velocity = u m/s

Acceleration = -g   ....(Since gravity is acting downwards)

Final velocity = 0 m/s    ........(As when the object reaches top the velocity is 0)

Let the time taken by the object be t seconds...

By Newton's 1st equation of Motion

v = u + at

0 = u + (-g)t    .............(Since at topmost point the velocity is 0)

u = gt

t= u/g  ....(1)

Now By Newton's 2nd equation of Motion

s = ut + 1/2 at^2

s = u*u/g  - 1/2 g*(u/g)^2   ..........(From 1)

s = (u^2/g)   -  1/2 (u^2/g)

s = u^2/2g

therefore the distance travelled by the body projected upwards is u^2/2g.

where the distance travelled by the object is not depended on the time taken !!!

Hope this Helps you !!