Question

Question #8efea

Answer 1

The ball is `(3h)/4` meters from the ground at `t = T/2`

Explanation:

I'm neglecting air resistance, but can add it in upon request. The body moves in only one dimension, which we shall call `y`. We shall define the origin of our coordinate system at the ground (`y=0`) and the initial position of the body `y(0) = h`. As is the usual convention we define upwards as positive.

Without air resistance, the only force acting on the body will be gravity - this acts in the negative y direction to provide an unbalanced force which we can use Newton's second law on:

`m(d^2y)/(dt^2) = -mg`

`(d^2y)/(dt^2)` represents acceleration in the y direction.

If you are unfamiliar with using calculus to solve problems like this then skip to the blue heading, this is just the derivation of a formula common to introductory dynamics.

`therefore (d^2y)/(dt^2) = -g`

Integrating wrt t gives:

`(dy)/(dt) = -g t + C`

The body starts from rest so `y'(0) = 0`.

`implies C =0`

`(dy)/(dt) = -g t`

Integrate again wrt t:

`y(t) = -1/2g t^2 + C`

`y(0)= h = C`

`implies y(t) = h - 1/2g t^2`

This is our expression for the displacement from the origin as a function of time. Notice it's very similar to the common expression

`s = v_0t + 1/2at^2`

which describes displacement under constant acceleration. This is because it is the exact same derivation, we have just accounted for the fact that initial speed is zero while doing it.

`color(blue)("Skip To Here")`

Anyway, onto solving the problem:

`y(t) = h - 1/2g t^2`

If it takes T seconds to reach the ground, `y(T) = 0`.

`y(T) = 0 = h - 1/2gT^2`

`implies h = 1/2gT^2`

`T^2 = (2h)/g`

`T = sqrt((2h)/g)`

Note that we discard the negative solution from the square root because negative time doesn't really make sense.

`T/2 = 1/2sqrt((2h)/g) = sqrt(h/(2g))`

`y(T/2) = h - 1/2g (sqrt(h/(2g)))^2`

`y(T/2) = h - 1/2g(h/(2g)) = h - h/4 = (3h)/4`

Answer 2

The body is at `h/4 m` from the top of tower.

Explanation:

Kinematic equation for the body dropped from a height `h` is
`h=ut+1/2g t^2`
where `u,g and t` are initial velocity, acceleration due to gravity and time taken to reach ground respectively.
Inserting given values we get
`h=0xxT+1/2g T^2`
`=>h=1/2g T^2` .......(1)

Similarly height `h_1` dropped in time `T/2` is
`h_1=1/2g (T/2)^2`
`=>h_1=1/8g T^2`......(2)
Dividing (2) by (1) we get
`h_1/h=1/4`

`=>h_1=h/4`