What would t= be ?

Question

An object is thrown upwards at time t = 0. The object's helght, h in feet, seconds later is given by the formula h = - 16t ^ 2 + 2t + 6 Determine the number of seconds t required for the object to return to the ground (h = 0)

Answer 1

h(t) =-16t^2 + 2t + 6 set h=0 for ground level

0 = -16t^2 + 2t + 6 multiply by -1

16t^2 -2t -6 = 0 divide by 2

8t^2 - t - 3 = 0 use the quadratic formula, t = -b/2a + or - (1/2a)sqr(b^2 -4ac) a = 8, b=-1, c=-3

t = 1/16 + or - (1/16)sqr(1+32(3)) = (1+ or - sqr96)/16 = about about 0.678 seconds to reach the ground = about 2/3 of a second

-16t^2 is the deceleration due to gravity, -32 feet per second per second

2 = initial velocity = 2 ft per second

maximum height is when the derivative, h'(t) = 0 = -32t +2 or t = 1/16 of a second

6 feet = initial height

max height = -16(1/16)^2 + 2(1/16) + 6 = 6.0625 feet.

it barely rises at all, then spends 2/3 of a second going down to h=0

Answer 2

Hi Zam A

Thanks to Vasile P above who left the perfect set up for now just use the Quadratic Formula.

Give it a try.