For this case we have the following equation:
h (t) = - 12t2 + 36t
When the object hits the ground we have:
- 12t2 + 36t = 0
We look for the roots of the polynomial:
t1 = 0
t2 = 3
Therefore, the time it takes the object to hit the ground is:
t = 3 s
Answer:
the time when the object hits the ground is:
t = 3 s
Answer: 2 - 2*sin³(θ) - √1 -sin²(θ)
Step-by-step explanation: In the expression
cos(theta)*sin2(theta) − cos(theta)
sin (2θ) = 2 sin(θ)*cos(θ) ⇒ cos(θ)*2sin(θ)cos(θ) - cos(θ)
2cos²(θ)sin(θ) - cos(θ) if we use cos²(θ) = 1-sin²(θ)
2 [ (1 - sin²(θ))*sin(θ)] - cos(θ)
2 - 2sin²(θ)sin(θ) - cos(θ) ⇒ 2-2sin³(θ)-cos(θ) ; cos(θ) = √1 -sin²(θ)
2 - 2*sin³(θ) - √1 -sin²(θ)
Answer:
x=−50
Step-by-step explanation:
Answer:
35
Step-by-step explanation:
40=1/2(115-x)
80=115-x
-35=-x
x=35
