a ball is kicked in the air from the top of a cliff. the path the ball travels is given by the equation h(t)=-5t^2+17t+22 where h(t) is the height in metres and t is the time in seconds

Question

a) what is the maximum height that the ball reaches
b)when will the ball hit the ground

Answer 1

Since the coefficient of t^2 is negative, the parabola opens downward and the max. point on the curve is the
vertex or max. ht.

V(h,k).
a. h = Xv = -b / 2a = -17 / -10 = 1.7.
Substitute 1.7 for t in the given Eq
and get:

h = k = Yv = 36.45m. = Height above cliff.

V(1.7,36.45).

b. H = 22 + 36.45 = 58.45m. = Ht. above
ground.

t(up) = (Vf - Vo) / g,
t(up) = (0 - 17) / -9.8 = 1.73s.

h = Vo*t + 0.5g*t^2 = 58.45m,
17t + 4.9*t^2 = 58.45,
4.9t^2 + 17t - 58.45 = 0,
Use Quadratic Formula to find t:
t(dn) = 2.13s.

T = t(up) + t(dn) = 1.73 + 2.13 = 3.86s
= Time in flight=Time to reach ground.

Answer 2

Correction:
h = Vo*t + 0.5*t^2 = 58.45m,
0 + 4.9t^2 = 58.45,
t^2 = 11.93,
t(dn) = 3.45s.

T = t(up) + t(dn) = 1.73 + 3.45 = 5.18s. = Time in flight = Time to reach ground.