A ball is released out a window that is 13 m above the ground. How fast will the ball be traveling when it reaches the ground? A. 32 m/s

B. 0.26 km/s

C. 0 m/s

D. 16 m/s

Option D. 16 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Height (h) = 13 m

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) =?

We can determine how fast (i.e final velocity) the ball is traveling when it reaches the ground as illustrated below:

v² = u² – 2gh

v² = 0² – (2 × 9.8 × 13)

v² = 0 – 254.8

v² = 254.8

Take the square root of both side.

v = √254.8

v = 15.96

v ≈ 16 m/s

Thus, the ball is travelling with a velocity of 16 m/s when it reaches the ground

the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft