Raymond B. answered • 01/18/22
Math, microeconomics or criminal justice
-4.9t^2 +12.4t + 2.9 = 1.8 (assuming there's a typo and the problem meant 12.4t not 12.4)
-4.9t^2 + 12.4t + 1.1 = 0
use the quadratic formula t= -b/2a + or - (1/2a)sqr(b^2 -4ac) (ignore the negative square root
t = -12.4/-9.8 + (1/9.8)sqr(12.4^2 + 4(4.9)(1.1))
t = 62/49 + sqr(153.76 + 21.56)/9.8 =1.265 + 1.351= 2.616
t = approximately 2.616 seconds to reach 1.8 meters, rounded off to 3 decimal places
check the answer
plug t=2.616 into the height equation: h(t) = -4.9t^2 + 12.4t + 2.9
-4.9(2.616) ^2 + 12.4(2.616) + 2.9 = -33.533 + 32.438 + 2.9 = 1.0805
the h(t) equation graphically is a parabola with initial height 1.1 meters, initial velocity 12.4 m/sec, with -4.9 = -9.8/2 which mean -9.8 meters per second per second is the deceleration, due to gravity
velocity = h'(t) = -9.8t + 12.4. set = 0 to solve for when maximum height is reached
t = 12.4/9.8 = 62/49 = 1.265 seconds to reach max height
h(1.265) = -4.9(1.265)^2 + 12.4(1.265) + 1.1 = -7.841 + 15,686 + 2.9 = 10.745 meters=max h
(1.265, 10.745) is the vertex of the parabola or about (1.3, 10.7)
(2.616, 1.8) is another point on the parabola or about (2.6, 1.8) roughly twice the time to reach max height
UNLESS you really meant h = -4.9t^2 + 12.4 + 2.9, with no t term
then initial velocity = 0, and initial height = 15.3 meters. with the ball not thrown but just dropped
h = -4.9t^2 +15.3 = 1.8
-4.9t^2 + 15.3-1.8 = 0
-4.9t^2 + 13.5 = 0
4.9t^2 = 13.5
t^2 = 13.5/4.9
t = square root of 13.5/4.9 = 1.660 seconds = about 1.7 seconds to reach 1.1 meters height
which is virtually the same as the first solution of 1.616 seconds
