Posted by gagan on .
A water balloon is shot into the air so that its height h, in metres, after t seconds is
h = —4.9t^2 + 27t + 2.4
a)How high is the balloon after 1 s?
b)For how long is the balloon more than 30 m high?
c)What is the maximum height reached by the balloon?
d)When will the balloon hit the ground?
Can you please help me thanks and need this urgently plz plz plz plz and thanks a lot for doing this.
a) just sub t=1 into the function
b) set -4.9t^2 + 27t + 2.4 = 30
and solve. You should get two solution for t
The balloon will be above 30 m for all values of t between those two solutions
c) find the vertex, the max height is the h coordinate of the vertex
Use this method:
for y = ax^2 + bx + c, the x of the vertex is -b/(2a)
once you have that value, sub it back in to get the y.
(of course you are dealing with t instead of x, and h instead of y)
d) set —4.9t^2 + 27t + 2.4 = 0 and solve for t
One of the values of t will be negative, that would be an extraneous root.
thnx a lot now i fully understand