object is thrown off a cliff initial speed of 4 feet per sec, and its speed in ft/sec after t seconds is given by s(t)=32t+4.

Question

If the object lands after 8 seconds, how high (in ft) is the cliff?

Answer 1

normally speed is v(t) not s(t) just to note

here distance(t) = d(t)=∫(32t+4)dt evaluated from t=0 to t=8

d=16t²+4t evaluated t=0 to t=8

d=1056 ft.

Answer 2

Plug in t=8:

s(t) = 32t+4

s(8) = 32(8)+4 = 256+4 = 260

Thus, the cliff is 260ft high.

Hope this helped!