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
I think it could be any 2 in between because it says greater than Luke's decimal which is .20 and less than Bekkas decimal which is .30
The second one because of the points and it would need to be pc=13
12 girls for every 5 boys
boys is second so it shouldb be indenomenaotr (bottom)
girls/boys
12girls/5boys=2.4girls/boy
answer is B
