The function of the object height is an illustration of a projectile motion
The object will never hit the ground
<h3>How to determine when the object hits the ground?</h3>
The function is given as:
h=16t^2+80t+96.
When the object hits the ground, h = 0
So, we have:
16t^2+80t+96 = 0
Divide through by 16
t^2+5t+6 = 0
Expand
t^2 + 3t + 2t + 6 = 0
Factorize
t(t + 3) + 2(t + 3) = 0
Factor out t + 3
(t + 2)(t + 3) = 0
Solve for t
t = -2 or t = -3
The time (t) cannot be negative.
Hence, the object will never hit the ground
Read more about projectile motion at:
brainly.com/question/1130127
Answer: 
Step-by-step explanation:
Given
The tower is at a height of 
the angle of depression is 
Suppose the surfer is x ft away from the base of the tower

Therefore, the surfer is at a distance of 19.31 ft from the base of the tower.
Answer:pls is it all we are to ans
Step-by-step explanation:
Answer:
Step-by-step explanation:

Answer:
x = $0.50
y= $0.75
Step-by-step explanation:
1. Multiply the equations to have the same coefficients
5(6x + 6y = 7.5) → 30x + 30y = 37.5
3(10x + 5y = 8.75) → 30x + 15y = 26.25
2. Subtract the equations
30x + 30y = 37.5
<u>- 30x + 15y = 26.25</u>
15y = 11.25
3. Solve for y by dividing both sides by 15
y = 0.75
4. Plug in 0.75 for y into one of the equations
6x + 6(0.75) = 7.5
5. Simplify
6x + 4.5 = 7.5
6. Solve for x
6x = 3
x = 0.5