Question

A particle is projected vertically upward and its height X in metre above the point of projection is given by X equal to 9.81 bracket start 40 minus half t square find its velocity and acceleration at I equal to 3 seconds and second equal to 5 seconds​

Answer 1

a particle is projected vertically upwards...

a particle is projected vertically upwards and it reach the Maximum height H in time T sec. the height of the particle at any time t will be

Answers :

As the height is not depends on time , It ol depend on initial velocity and gravity

As we ol know that the maximum height ,

Hmax = u2/2g

So, at any time t , the maximum height (Hmaz) = u2/2g

one year ago

We know that a certain height is reached two times in a vertical projection.(One while going upward,one while going back downward)

let the height reached in time t be h

maximum height=H

difference in height=H-h

now H is reached in T and h is reached in t time.So H-h is reached in (T-t) time ...(1)

while going back down H-h is reached in (T-t)seconds

So,

H-h=u(T-t) + ½g(T-t)2

but u =0 (velocity at max height is zero)

So,

H-h = ½g(T-t)2

h = H – ½g(T-t)2

i hope it will help u....
please mark it as brainliest...
#FOLLOW ME