Physics: Arock is thrown vertically upward from ground level at time t = 0. at t = 1.6 s it passes the top of a tall tower, and 1.0 s later it reaches its maximum height. what is the height of the tower?

Arock is thrown vertically upward from ground

Turtlelover05

10.03.2022

The rock has height at time according to

$y=v_0t-\dfrac g2t^2$

where v_0 is the velocity with which it was thrown, and g = 9.8 m/s^2 is the acceleration due to gravity.

Complete the square to get

$y=\dfrac{{v_0}^2}{2g}-\dfrac g2\left(t-\dfrac{v_0}g\right)^2$

which indicates a maximum height of $\dfrac{{v_0}^2}{2g}$ occurs when $t=\dfrac{v_0}g$ . We're told this time is 2.6 s after the rock is thrown:

$2.6\,\mathrm s=\dfrac{v_0}{9.8\frac{\rm m}{\mathrm s^2}}\implies v_0=25.48\dfrac{\rm m}{\rm s}$

So when t = 1.6 s, the rock reaches the tower's height of

$y=v_0(1.6\,\mathrm s)-\dfrac g2(1.6\,\mathrm s)^2\approx\boxed{28\,\mathrm m}$

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emaleyhughes21

08.04.2022

Dot structures do not show the distribution of electrons in orbitals and take up a lot of space.

Arrow and line diagrams take up a lot of space and make it difficult to count electrons.

Written configurations make it easy to lose count of electrons and do not show the distribution of electrons in orbitals.

(re-write in your own words this is off engenuity)

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