A ball is thrown from a height of 70 feet with an initial downward velocity of 4 ft/s. The ball's height h (in feet) after t seconds is given by the following. h=70-4t-16t^2.

Question

How long after the ball is thrown does it hit the ground?

Round your answer(s) to the nearest hundredth.

Answer 1

h(t) = 70 -4t -16t^2 = 0

use the quadratic formula, ignore the negative square root

t= -b/2a + (1/2a)sqr(b^2 -4ac)

t = 4/-32 + (1/32)sqr(4^2 +280(16))

= -1/8 + (sqr281)/8

=about (-1+17)/8

= about 16/8

= about 2.0 seconds to reach the ground

in 1 second h=70-4-16 = 50 feet off the ground

in 2 seconds h= 70-4(2)-16(2^2) = -2 feet underground

slightly less than 2 seconds it hits the ground

in 1.7 seconds h = 70-4(1.7)-16(1.7^2) = 17 feet above ground

in 1.8 seconds h= 70-4(1.8)-16(1.8) = 11 feet above ground

in 1.9 seconds h = 70 -4(1.9)-16(1.9) = 5 feet above ground