<span>going downstream -
t = 14/(2 + v)
going upstream -
t = 7/(2 - v)
14/(2 + v) = 7/(2 - v)
(2 + v)/2 = 2 - v
v = 2/3km/h</span><span>
</span>
Answer:
162
Step-by-step explanation:
( -a ) ( b ) ( -a + b )
-6 ( 3 ) ( -6 + -3 )
-6 ( 3 ) ( -9 )
- 18 ( -9 )
162
In order to find height from where ball is dropped, you have to find height or h(t) when time or t is zero.So plug in t=0 into your quadratic equation:h(0) = -16.1(0^2) + 150h(0) = 0 +150h(0) = 150 ft is the height from where ball is dropped. When ball hits the ground, the height is zero. So plug in h(t) = 0 and solve for t.0 = -16.1t^2 + 15016.1 t^2 = 150t^2 = 150/16.1t = sqrt(150/16.1)t = ± 3.05Since time cannot be negative, your answer is positive solution i.e. t = 3.05
