You throw a ball downward from a window at a speed of 2.0 m/s. How fast will it be moving when it hits the sidewalk 2.5 m below?

Objects thrown downward has a constant acceleration (due to gravity) which is approximately equal to 9.8 m/s². The given can be generally related to each other using the quation,

2ad = (Vf) ² - (Vi) ²

where a is the acceleration (9.8 m/s²), d is distance (2.5 m), Vf is the unknown final velocity, and Vi is initial velocity (2 m/s).

Substituting the given,

2 (9.8 m/s²) (2.5 m) = (Vf) ² - (2 m/s) ²; Vf = 7.28 m/s

Thus, the final velocity is 7.28 m/s.

This will require the use of kinematic equations:we have:Vstarting = 2.0 m/sHeightStarting = 2.5 mand we need to find Vfinal

the kinematic equation of V^2 = V^2 + 2ah can be used velocity final ^2 = velocity starting ^2 + 2 * acceleration of gravity * change of height

there are two terms which we are given, velocity starting and change in height, we are missing two terms: final velocity (which we are looking for) and acceleration of gravity.

acceleration of gravity is a number which is always the same, and can be found usually in the front or back cover of a physics book. we are using meters, so we will use 9.81 m/sec^2

so now we have 3 knowns, and one unknown.

all that is left is to plug the three known values into the kinematic equation I listed above, and use algebra to solve for the one unknown, which is final velocity.

i guess it would be better to uses=ut+1/2at2

and find t

then use v=u+at

where s is the distance=2.5

u=initial speed=0

v=final speed which we have to find

a=acceleration which is due to gravity (take as positive)

t=time

v = square root (v^2+2ah)

v = square root (2^2 + 2∗9.8∗2.5)

v = square root (53)

v = 7.3