Question

The height of a ball tossed vertically up from 15 m with speed of 10m/s described by
h(t) =15+10t-5t time in sec after ball thrown and h height above ground
How many sec before ball falls to ground
What is max height
Pls help cheers

Answer 1

h(t) =15 + 10t - 5t
should one of these be t^2 ...?

Answer 2

if its-5t^2+10t+15, then the ball will be in the air for 3 seconds and reach a max height of 20

Answer 3

yes one is 5t squared

Answer 4

Why did you put the 15 outside the brackets and where did the 4 come from? Cheers

Answer 5

at t=1 ud be at -5+10+15 = 20, vertex is at (1,20)

Answer 6

-5t^2+10t+15

Vertex: \(\ \sf x = \dfrac{-b}{2a} \)

a = -5
b = 10

\(\ \sf x = \dfrac{-10}{-10} \implies 1\)

Now substitute that into the function: h(t) = -5t^2+10t+15 and that'll be the "y" part of the coordinate.

you already have the "x," hence the "x = " part.

Therefore (1, ?) is the vertex ~

Finish it