The answer is -51, hope this helps
Answer:
469.42 ft²
Step-by-step explanation:
w/sin27 = 38/sin40
w = sin27*38/sin40
w = 26.84 ft
∡X = 180 - 27- 40 = 113º
A = 0.5*(26.84)*(38)*sin(113)
A = 469.42 ft²
The given equation is 
where h is the height, in feet, of a ball and t is the time, in seconds.
<u>Part a: The height of the ball when t = 2 seconds:</u>
The height of the ball above the ground 2 seconds after it is released can be determined by substituting t= 2 in the equation , we get;
Simplifying the terms, we get;
Thus, the height of the ball after 2 seconds is 100 feet.
<u>Part b: The height of the ball when t = 4 seconds:</u>
The height of the ball above the ground 4 seconds after it is released can be determined by substituting t = 4 in the equation , we get;
Simplifying the terms, we get;
Thus, the height of the ball after 4 seconds is 68 feet.
Answer:
maximum height is 4.058 metres
Time in air = 0.033 second
Step-by-step explanation:
Given that the equation height h
h = -212t^2 + 7t + 4
What is the toy's maximum height?
Let us assume that the equation is a perfect parabola
Time t at Maximum height will be
t = -b/2a
Where b = 7 and a = - 212
t = -7/ - 212 ×2
t = 7/ 424 = 0.0165s
Substitute t in the main equation
h = - 212(7/424)^2 + 7(7/424) + 4
h = - 0.05778 + 0.115567 + 4
h = 4.058 metres
Therefore the maximum height is 4.058 metres
How long is the toy in the air?
The object will go up and return to the ground.
At ground level, h = 0
-212t^2 + 7t + 4 = 0
212t^2 - 7t - 4 = 0
You can factorize the above equation and pick the positive time t since time can't be negative
Or
Since we have assumed that it's a perfect parabola,
Total time in air = (-b/2a) × 2
Time in air = 0.0165 × 2 = 0.033 s
