a) 70.1 m
The ball is moving by uniformly accelerated motion, with constant acceleration 
(acceleration due to gravity) towards the ground. The height of the ball at time t is given by the equation
where 
is the height of the ball at time t=0. Substituting t=1 s, we can find the height of the ball 1 seconds after it has been dropped:
b) 3.9 s
We can still use the same equation we used in the previous part of the problem:
This time, we want to find the time t at which the ball hits the ground, which means the time t at which h(t)=0. So we have
And solving for t we find
If the slope is 6/11 then, in y=mx+b form, m will be 6/11. When you multiply 6/11 and -1 you get -6/11. To get the y (-6), you have to add -60/11. So in, y=mx+b form the equation is y=6/11x+(-60/11). To make it standard form, you add negative 6/11x to both sides and then multiply by 11 to get rid of the fractions. So therefore the equation would be:
-6x + 11y = - 60
Idk is I’m right but....25%x4=1 which is the same as 100%
Answer:
y = 2x -2
Step-by-step explanation:
This is only the answer if you need the slope-intercept form equation.
