A ball is thrown into the air with a initial velocity of 25 meters per second. The function h(t)+-4.9t^2+25t+6 represents the height of the t in seconds. Ball above the ground in meters, with respect to time. What would be a reasonable domain for 36.9 meters

[2.1, 3]

Step-by-step explanation:

Given the height of the ball modelled by the equation;

h(t) =-4.9t^2+25t+6

Required

Domain for 36.9m

To get that we are to find the value of the input variable which is the time t  when h(t) = 36.9

Substitute

36.9 = -4.9t^2+25t+6

multiply through by -10

-369 = 49t²-250t-60

49t²-250t-60 + 369 = 0

49t²-250t+309 = 0

Find the domain of t

49t²-147t-103t+309 = 0

49t(t-3)-103(t-3) = 0

(49t-103)(t-3) = 0

49t - 103 = 0 and t-3 = 0

49t = 103 and t = 3

t = 103/49 and t = 3

t = 2.1 and t = 3

Hence the reasonable domain for 36.9m is [2.1, 3]

6 on both sides and twelve on the top and the bottom

step-by-step explanation:

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°