- 8

- 0

trying to study for a midterm with kinematics 1d 2d and dynamics

Heres problem:

A stone is thrown vertically downward from the top of a 96m building at a velocity of 5.0m/s. This stone is observed passing by a window that has a height of 2.0m. Assuming that the bottom of the window is 25m from the ground, how long (time) does it take for the stone to pass the window?

-------

Heres what Ive done so far:

distance = d = 96m

time= t = ?

constant velocity = v = 5.0m/s

acceleration = a = 9.81m/s^2

so i used v=d/t to find total time (5)=(96)/(t) I got 19.2 seconds.

So to find the time it takes to pass the window i was thinking of using a kinematics equation. Because we know the window is 25m from the ground we know that during the last 25m of free fall that velocity initial = vi = 5.0m/s and velocity final = vf = 0 m/s

OK so i think the above time is wrong and using a kinematics equation i found a different total using gravity, distance etc.. I got 4.424 seconds. well Im sorta lost... anyways I was pugging in a bunch of numbers into different formula.. I found the bottom 25m and the 69m above the 2 m window and tried a subtraction... but got wrong answer... it was negative.... well could some one show me how to do this question the right way..

BTW.. the real answer is: (5.3x10^-2 sec)

Thanks for help

Heres problem:

A stone is thrown vertically downward from the top of a 96m building at a velocity of 5.0m/s. This stone is observed passing by a window that has a height of 2.0m. Assuming that the bottom of the window is 25m from the ground, how long (time) does it take for the stone to pass the window?

-------

Heres what Ive done so far:

distance = d = 96m

time= t = ?

constant velocity = v = 5.0m/s

acceleration = a = 9.81m/s^2

so i used v=d/t to find total time (5)=(96)/(t) I got 19.2 seconds.

So to find the time it takes to pass the window i was thinking of using a kinematics equation. Because we know the window is 25m from the ground we know that during the last 25m of free fall that velocity initial = vi = 5.0m/s and velocity final = vf = 0 m/s

OK so i think the above time is wrong and using a kinematics equation i found a different total using gravity, distance etc.. I got 4.424 seconds. well Im sorta lost... anyways I was pugging in a bunch of numbers into different formula.. I found the bottom 25m and the 69m above the 2 m window and tried a subtraction... but got wrong answer... it was negative.... well could some one show me how to do this question the right way..

BTW.. the real answer is: (5.3x10^-2 sec)

Thanks for help

Last edited: