Answer:
So the given equation is h= -5t^2+40t
the first question asks when it will be 75 meters above the ground. So plug in 75 for h and find t
75= -5t^2+40t
start by dividing everything by 5
15= -x^2 + 8x
then you can add x squared to both sides and subtract 8x
x^2 - 8x + 15 =0
then you can plug it into the quadratic formula, which is a pain to type so I did it on paper.
x = 5 or 3
so it will be 75 meters above the ground at 3 seconds and 5 seconds
Rewriting the equation
ok so I accidentally already factored it when I divided everything by 5...
I believe the answer here is x^2 - 8x + 15 =0
The meaning of the x intercepts is when h=0 or the object is 0 meters above the ground (basically when it starts and when it ends)
the maximum height is greater than 75 because it gets to 75 meters 3 seconds after it is thrown, and continues going higher before it comes back to 75 two seconds later.
The object is airborne for 8 seconds because the second x intercept is at (8,0)
