Question

An object is launched at 19.6 m/s from a height of 58.8 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t^2 +19.6t + 58.8. What is the object's maximum height?
A. 58.8
B. 194.04
C. 78.4
D. 117.6

Answer 1

differentiate
and put derivative equal to 0
it will give
-9.8t+19.6=0
t=2
that is the time when it reaches max hieght now plug that into eq of h(t)
u will get the answer

Answer 2

78.4

Answer 3

did u get that

Answer 4

@fionakm