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
Answer:
SA = 384 in ^2
Step-by-step explanation:
The surface area of a cube is given by
SA = 6 S ^2 where s is the side length
SA = 6 * (8) ^2
SA = 6 * 64
SA = 384 in ^2
Answer:
A. 13
Step-by-step explanation:
Answer:
I do not believe you have inserted all the information needed to solve this problem.
Step-by-step explanation:
Answer:
4.16, 4.6, 4.608, 8.320
Step-by-step explanation:
