Question #94a89

2 Answers
Nov 28, 2017

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)

Nov 28, 2017

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