Antiderivatives

Question

I need to find the answer to this question. I know that A is 717. But I cannot find the answer for B or C.

A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 12 feet per second. A. Remembering that the acceleration due to gravity is -32 ft per second squared, how high is the stone 4 seconds later? B. At what time does the stone hit the ground? C. What is the velocity of the stone when it hits the ground?

The answers that did not work:

B: -7.24/7.24

C: -219.68/234.09

Answer 1

h(t) =-16t^2 +12t +925

h(4) = -16(4^2) +12(4) +925

= -256 +48+925

= 973-256

= 717 feet high after 4 seconds

it hits the ground when h=0

h(t) = -16t^2 +12t +925

t = 12/32+(1/32)sqr(144+4(16)(925))

= (12+sqr59344)/32

= about 7.987695 seconds to hit the ground

= about 7.99 seconds rounded to 2 decimal places

= about 8 seconds

h(8)= -16(8)(8) +12(8)+925

=-116(8)+925

= -928 +925=-3

that extra .01 second put it 3 feet below ground

velocity = integral of h(t)

= -32t +12

in 8 seconds

velocit =-32(8)+12

= -256+12 = -244 ft/sec when it hits ground, rounded to nearest integer

=-32(12+sqr59344)/32 +12

= -12 -sqr59344 +12

=- sqr59344

= about - 243.606239

= about -243.61 ft/sec rounded to two decimals