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 minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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