Question

an object is dropped from height=h with some initial velocity. if it covers 1/4th of its distance in 1/2 time taken and 1/2 distance is covered in 6√2 second. find h.

Answer 1

S=Vit+12at2S=Vit+12at2 ————--eqn 1

Vf=Vi+atVf=Vi+at —————eqn 2

combine equation 1 and equation 2 to eliminate “t” gives

V2f−V2i=2aSVf2−Vi2=2aS —————eqn 3

It is highly recommended to watch your signs in these equations. Velocities are up = positive, down = negative and the acceleration due to gravity always points down so ay=−9.81m/s2

Consider the first half of the fall from point 1 to point 2. We know the distance is h/2 and the time to fall is (t-1), so let’s use equation 1 written from point 1 to point 2. I’ll add subscripts since we are writing the equation in the y-direction:

Sy=(Vi)yt+12ayt2Sy=(Vi)yt+12ayt2

Watching our sign convention, the position (at point 2) is below the starting point (at point 1), therefore Sy=−S1=−h2Sy=−S1=−h2

−h2=V1(t−1)+12(−9.81)(t−1)2−h2=V1(t−1)+12(−9.81)(t−1)2

but V1=0V1=0

−h2=+12(−9.81)(t−1)2−h2=+12(−9.81)(t−1)2

or

h=9.81(t−1)2h=9.81(t−1)2 ————- equation 4

Now let’s write equation 1 again for the entire fall from point 1 to the ground.

Sy=(Vi)yt+12ayt2Sy=(Vi)yt+12ayt2

The ground is a distance hh below the starting point (at point 1), therefore Sy=−hSy=−h

−h=V1(t)+12(−9.81)(t)2−h=V1(t)+12(−9.81)(t)2

and again V1=0V1=0

−h=−4.905(t)2−h=−4.905(t)2

or

h=4.905(t)2h=4.905(t)2————- equation 5

solving equations 4 and 5 gives:

9.81(t−1)2=4.905(t)29.81(t−1)2=4.905(t)2

t2−2t+1=12t2t2−2t+1=12t2

12t2−2t+1=012t2−2t+1=0

or

t2−4t+2=0t2−4t+2=0

t=−(−4)±(−4)2−4(1)(2)√2(1)t=−(−4)±(−4)2−4(1)(2)2(1)

or

t=4±(16−8)√2t=4±(16−8)2

t=3.41sect=3.41sec

or

t=0.586sect=0.586sec

EDIT: But the question implied that there was more than 1 second of flight time, so t=0.586 sec is invalid. Thanks to Jake for pointing out my oops.

Final answer: t=3.41sect=3.41sec. Use equation 4 or 5 to determine the height.