What is the ejection speed of a ball if it was shot horizontally 1.1 meter above the floor, and touched it 2.44 meters further from the vertical projection of the ejection point? use 9.810 m/s^2 for the acceleration of gravity.

5.1 m/s

Explanation:

First we consider the vertical motion of the ball, which is a uniform accelerated motion (free fall). So can use the suvat equation

:

where

, choosing downward as positive direction:

s = 1.1 m is the vertical displacement

u = 0 is the initial vertical velocity of the ball

t is the time

is the acceleration of gravity

Solving for t, we find the time at which the ball hits the ground:

Now we consider the horizontal motion. This is a uniform motion with constant horizontal velocity. Therefore, the horizontal distance travelled by the ball is given by

where

d = 2.44 m is the horizontal distance travelled (we are told that the ball hits the ground 2.44 m further from the vertical projection of the ejection point)

t = 0.474 s is the time of flight

Solving for we find

The horizontal velocity is constant during the whole motion (since there are no forces acting along this direction), so this is the ejection speed of the ball.

the body's situation won't change the forces are equal opposit and on the same body

d eras and periods