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
Answer:
and is in photo given.I didn't get time to type.
Answer:
I think he would be dead poggers
Explanation:
V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)
One km^3 is 1,000,000,000 m^3=10^9 m^3 hence 3.73 10^8 km^3 is 3.73 10^17 m^3
One meter is 3.28084 feet hence 1 m^3 is (3.28084)^3 feet
Thus 3.73 10^8 km^3 is 3.73*35.315 10^17 = 132 cubic feet
