Question

Question #879f1

Answer 1

The body will be `60m` above the ground at `2` seconds after being released.

Explanation:

Assuming that the acceleration due to gravity has been given to be `g=-10m//s^2` we can verify that an object with zero initial velocity, `v_o=0m/s`, would undergoing constant acceleration would move `80m` in `4` seconds using the equation of motion

`x(t) = 1/2 a*t^2 + v_o * t + x_o`

which gives us

`h(t) = 1/2 g*t^2 + x_o = -1/2 10 t^2 + 80 = -5t^2 + 80`

where we have chosen the position to be the height, and the acceleration must be negative, since it tries to decrease the height of objects.

At 4 seconds this gives us:

`h(4) = -5*4^2 + 80 = -80 + 80 = 0`

which implies that it has hit the ground as expected. Also, at `t = 0` we get

`h(0) = -5*0^2 + 80 = 80`

as expected from the question. Now that we are confident that we have the right equations, we can use it to find the height at `2` seconds:

`h(2) = -5*2^2 + 80 = -20 + 80 = 60`

The body will be `60m` above the ground at `2` seconds after being released.

NOTE

It's important to note that I'm assuming that the acceleration due to gravity is `10m//s^2` rather than the usual `9.8m//s^2` since that is the only way for a body being released at an initial velocity of `0m//s` to hit the ground in 4 second. If I do not make this assumption there is an initial velocity to solve for which I think is more complex than the question is intended to be.

Answer 2

`sf(60color(white)(x)m)`

Explanation:

Use `sf(s=1/2"g"t^2)`

`sf(s=1/2xx9.8xx2^2=19.62color(white)(x)m)`

This is the distance the body has fallen so the distance above the ground is given by:

`sf(h=80-19.62=60.38color(white)m)`

`sf(=60color(white)(x)m` to 1 sig. fig.

You don't need to be given the height of the tower. You can use the same expression putting t = 4.