Question 1173179
A ball is thrown vertically upwards from ground level with a velocity of 28 m/s
use the formula: h = -4.9t^2 + 28t
(i) What was its maximum height above the ground?
max height occurs on the axis of symmetry. Find this using t = -b/(2a)
t= {{{(-28)/(2*-4.9)}}}
t =  2.857 seconds to reach max height
Find the height at this time
h = -4.9*2.857^2 + 28(2.857)
h = -40 + 80
h = 40 ft max height
:
(ii) How long does it take to reach 20 m height?
-4.9t^2 + 28t = 20
-4.9t^2 + 28t - 20 = 0, a quadratic equation, use the quadratic formula
a=-4.9, b=28, c=-20
two solutions
t = .8368 sec on the way up
and
t = 4.8774 sec on the way down