Problema Solution
If a metal ball is thrown downward with an initial velocity of 22 feet per second (15 mph) from a 100 foot water tower its height h in feet above the ground after t seconds. Determine symbolically when the geight of the ball is 62 feet.
Answer provided by our tutors
h(t)=-16t^2-22t+100
62=-16t^2-22t+100
-16t^2-22t+38=0
-8t^2-11t+19=0
solve by following quadratic formula
..
a=-8, b=-11, c=19
t=[-(-11)+-sqrt(-11)^2-4*-8*19)]2*-8
t=[11+-sqrt(729)]/16=(11+-27)/-16
t=-16/-16=1
t=38/-16=-2.375 (reject, t=>0)
ans:
The height of the ball is 62 feet above the ground after 1 second