Answer:
Step-by-step explanation:
First, the acceleration of gravity is -9.8m/s^2. This still works out since the formula uses 1/2 of it.
Hopefully you see where the other parts come from. Anyway, standard form of a quadratic is ax^2+bx+c=0. So this is almost there. You just need to subtract that 2.1 from both sides.
-4.9t^2+7.5t+-.3=0
Now with this, since it has taken into account the height that the ball was caught with that 2.1, you just need to find the 0s, which is what the quadratic equation does.
The quadratic equation is (-b±sqrt(b^2-4ac))/(2a) and we have a = -4.9, b = 7.5 and c = -.3. Remember you want to keep the signs. Now we just plug in.
(-b±sqrt(b^2-4ac))/(2a)
(-7.5±sqrt((-7.5)^2-4*-4.9*-.3))/(2*-4.9)
(-7.5±sqrt(56.25-5.88))/(.9.8)
(-7.5±sqrt(50.37))/(-9.8)
(-7.5±7.097)/(-9.8)
The plus or minus means there are two equations.
(-7.5+7.097)/(-9.8) and (-7.5-7.097)/(-9.8) So we will solve for both of these.
.04112 and 1.4895. That means these two times are when the ball is at 2.1 meters. One time on the way up and one time on the way down. We can safely assume that the other player catches the ball on the way down, so we want to use the second time, so 1.4895 seconds.
Answer:
45
Step-by-step explanation:
Another basic one
(1/2)^2*1/2*(1/2)^3=
(1/2)^6 =
1\2^6=1/64
if in decimal: 0.015 but i would prefer 1/64 if i was the teacher :)
Answer: could you please attach a photo no way anyone can do it without a photo
