A body thrown vertically upwards or vertically downwards or dropped from a height will move in a straight vertical line.
lf air resistance is ignored, the body will be subjected to acceleration due to gravitational force exerted by the earth, which is denoted by `g`. The value of `g` on the earth is `9.8m//s^2` in the downward direction.
For small heights, the value of `g` is constant, we can use equations of uniformly accelerated motion.
We shall take upward direction as positive & down direction as negative, as our convention.
`text(Motion of a particle projected downwards from height)` `h` `text(above surface of earth :)`
Suppose a particle is projected downwards from height `h` above the surface of the earth with speed u. To find the time taken by it to strike the surface of the earth, taking upward direction as positive, `u=-u, a= - g, s = -h` , apply `s= ut+1/2 at^2` , solve the quadratic and get the positive value of `t`.
`text(Motion of a particle dropped from a height)` `h` `text(above surface of earth :)`
Solve using `v^2 = u^2+2as` and `s=ut+1/2 at^2`, taking `u = 0`, Velocity with which it strikes the surface will be `sqrt(2gh)` and the time it will take to strike the surface will be `sqrt((2h)/g)`
A body thrown vertically upwards or vertically downwards or dropped from a height will move in a straight vertical line.
lf air resistance is ignored, the body will be subjected to acceleration due to gravitational force exerted by the earth, which is denoted by `g`. The value of `g` on the earth is `9.8m//s^2` in the downward direction.
For small heights, the value of `g` is constant, we can use equations of uniformly accelerated motion.
We shall take upward direction as positive & down direction as negative, as our convention.
`text(Motion of a particle projected downwards from height)` `h` `text(above surface of earth :)`
Suppose a particle is projected downwards from height `h` above the surface of the earth with speed u. To find the time taken by it to strike the surface of the earth, taking upward direction as positive, `u=-u, a= - g, s = -h` , apply `s= ut+1/2 at^2` , solve the quadratic and get the positive value of `t`.
`text(Motion of a particle dropped from a height)` `h` `text(above surface of earth :)`
Solve using `v^2 = u^2+2as` and `s=ut+1/2 at^2`, taking `u = 0`, Velocity with which it strikes the surface will be `sqrt(2gh)` and the time it will take to strike the surface will be `sqrt((2h)/g)`