Calculus I (Antiderivatives & Indefinite Integration): A stone is thrown straight up from the edge of a roof, 625 feet above the ground, at a speed of 10 feet per second.

Question

A stone is thrown straight up from the edge of a roof, 625 feet above the ground, at a speed of 10 feet per second.

A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 4 seconds later? I was able to correctly complete this step, with an answer of 409.

B. At what time does the stone hit the ground?

C. What is the velocity of the stone when it hits the ground? 

Answer 1

A. y(t) = 625 + 10t - 16t²; h = y(4) = 625 + 40 - 256 = 409 feet

B. y(t) = 625 + 10t - 16t² = 0; t = (- 5 ±√25 + 625·16)/(-16) = 6.57 s

C. v(t) = 10 - 32t ; v(6.57) = 10 - 32·6.57 = - 200.24 ft/s(directed down)

Answer 2

given a(t)=constant=-32 f/s²,,,,v(0)=10 f/s y(0)=625 f

v(t)=∫(-32)dt= -32t+C,,,,,t=0,,,,C=10

v(t)= -32t+10

y(t)=∫v(t)dt=∫(-32t+10)dt= -16t²+10t+K,,,,t=0,,,,K=625

y(t)= -16t²+10t+625

y(t)=0 ,,,

t=(10/32)+/-[√(10²+4*16*625)]/32

t=6.57 s

v(6.57)= -200.2 f/s= -143.3 mph