how to solve h=-16t^2+4t+48

Answer 1

h(t) =-16t^2 + 4t + 48

initial height = 48 feet

initial velocity = 4 feet per second

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

take the derivative to get velocity at time t

= v(t) = h'(t) = -32t +4

set it equal to zero to find the time of highest height

-32t + 4 = 0

t = 4/32 = 1/8 = 0.125 second

plug that into the original equation to find the maximum height

h(1/8) = -16(1/8)^2 + 4(1/8) + 48 = -1/4+1/2+48 = 48.25 ft

it goes up from 48 to 48.25 in an eighth of a second

then it goes down to the ground h=0 when

-16t^2 + 4t + 48 = 0

divide by -4

4t^2 -t - 12 = 0

t = 1/8 + (1/8)sqr(1+4(4)(12))

= 1/8 + about 13.892/8 = (0.125+13.892)/8 = 14.017/8 = 1.752 seconds to hit the ground