They could be 28 + 29 + 30
Hypotenuse = 5√6
Let the length of each leg be x, since it is Isosceles .
x² + x² = (5√6)²
2x² = (5 * 5 * √6 * √6)
2x² = 150
x² = 150/2
x² = 75
x = √75
x = √(25 * 3)
x = √25 * √3
x = 5√3
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
