Answer:
6.5 s
Step-by-step explanation:
The equation provided provides us with the height of the canonball at any time t. When the ball hits the ground, the height of the ball above the ground is zero.
Therefore
h(t)=-4.9t^2+30.5t+9.4 = 0
We then solve for t. Seeing as this is a quadratic equation, we can use the quadratic formula:
![t = \frac{-30.5-\sqrt{30.5^{2} -4(-4.9)(9.4)} }{2(-4.9)} = 6.5 s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-30.5-%5Csqrt%7B30.5%5E%7B2%7D%20-4%28-4.9%29%289.4%29%7D%20%7D%7B2%28-4.9%29%7D%20%20%3D%206.5%20s)
The quadratic formula gives two answers, a positive one and a negative one. The positive one is after time zero and the negative one is before time zero. We are interested in the one after time zero (time zero being the time when the canon is fired)
Answer:
D
Step-by-step explanation:
Answer:
the answer is 864mm's
Step-by-step explanation:
okay so this is my explanation for you
I got the answer 3f there you go
Answer:
combine the 2 like-terms (-52 and -12) move to the other side of the equals sign using opposite operations (add 64 to both sides) and divide both sides by 3
Step-by-step explanation: