Question

Question #94a89

Answer 1

OK, what we are facing is a kinematic question.

Explanation:

First job, list what you know in a neat table with units, put a question mark next to the thing you’re looking for and a dash next to the factors that are not mentioned:

s = ?m (displacement)
u = 15.9m/s (initial velocity)
v = 0m/s (final velocity, it stops at the highest point)
`a = - 9.81m/s^2`
(note the minus sign? It shows the acceleration vector is pointing in the opposite direction to the positive velocity vector)
t = -s

So we need one of the kinematic equations linking s, u, v and a.

There is only one: `v^2 = u^2 + 2a.s`

as `v = 0`, `v^2 = 0`

Rearranging gives: `s = -u^2/(2a`

`s = -15.9^2/(2xx-9.81) = 12.9 m` (to 3 s.f)

Answer 2

Use formulas `d=v_0t+1/2at^2` and `v_f=v_0+at`

Explanation:

A key understanding here is that the final velocity in the y-direction will be zero.

`v_0`=15.9 m/s and `v_f`= 0 m/s

First, calculate the time the stone will be in the air `v_f=v_0+at`

`0 = 15.9 m/s + -9.81 m/s^2*t`

`-15.9 m/s = -9.81m/s^2 t`

`t= {-15.9m/s}/-9.81m/s^2 = 1.62s`

Then, `d=v_0t+1/2at^2`

`d=15.9 m/s(1.62s)12+1/2(-9.81m/s^2)1.62s^2`

d = 12.88 m