Answer:
The velocity of the ball before it hits the ground is 381.2 m/s
Explanation:
Given;
time taken to reach the ground, t = 38.9 s
The height of fall is given by;
h = ¹/₂gt²
h = ¹/₂(9.8)(38.9)²
h = 7414.73 m
The velocity of the ball before it hits the ground is given as;
v² = u² + 2gh
where;
u is the initial velocity of the on the root = 0
v is the final velocity of the ball before it hits the ground
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 7414.73 )
v = 381.2 m/s
Therefore, the velocity of the ball before it hits the ground is 381.2 m/s
I think that numbers one, three, and four are true
Acceleration....................................... <span />
The uniform microwave radiation remaining from the Big Bang.
So, your body is always having background radiation and that means space!
Answer:
ax = -3.29[m/s²]
ay = -1.9[m/s²]
Explanation:
We must remember that acceleration is a vector and therefore has magnitude and direction.
In this case, it is accelerating downwards, therefore for a greater understanding we will make a diagram of said vector, this diagram is attached.
![a_{x}=-3.8*cos(30) = -3.29 [m/s^{2}]\\ a_{y}=-3.8*sin(30) = -1.9 [m/s^{2}]](https://tex.z-dn.net/?f=a_%7Bx%7D%3D-3.8%2Acos%2830%29%20%3D%20-3.29%20%5Bm%2Fs%5E%7B2%7D%5D%5C%5C%20a_%7By%7D%3D-3.8%2Asin%2830%29%20%3D%20-1.9%20%5Bm%2Fs%5E%7B2%7D%5D)