An object falls 16 t 2 16t2 feet in t t seconds. you drop a rock from a bridge that is 75 feet above the water. will the rock hit the water in 2 seconds? if so how many seconds explain

No it is still 11 feet from the water at 2 seconds

2.1651 seconds  is when it will hit the water

Step-by-step explanation:

h(t) = -16 t^2 +75

This is the formula  for the distance

The -16t^2 is due to gravity and the 75 is the height above the bridge

We want to know the height a t =2

h(2) = -16 (2)^2 +75

h(2) =-64 +75

h(2) = 11

The rock will be 11 ft above the water

When will the rock hit the water

h = 0

0 = -16 t^2 +75

Subtract 75 from each side

-75 = -16t^2 +75-75

-75 = -16t^2

Divide by -16

-75/-16 =  -16t^2 /-16

75/16 = t^2

Take the square root of each side.  We only take the positive square root since time must be positive

sqrt(75/16) =t

5/4 sqrt(3) = t

2.1651 seconds = t

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°