you have an object that you see free falling for time t for L meters, at what height was it dropped?

Question

the object was dropped some distance above you.
answer in form of a formula. L , g(gravity), t (the time you saw an object falling for a given length), and the height at which it was dropped (assuming that it stopped dropping after you saw it disappear)

Answer 1

L=V*t + 1/2*g*t^2 (thats t squared). V is any initial velocity (ie. you throw the object)

Answer 2

L/t = V, the average velocity for the time you saw it go through those L meters. This is also the instantaneous velocity at the point halfway through the L meters. Then you can plug that V into the formula

V^2 = Vo^2 +2gh

where Vo is initial velocity, apparently zero in this case
h is the height above the midpoint of the L meters.

Solve for h.

I'm not clear what this means "height at which it was dropped (assuming that it stopped dropping after you saw it disappear)". Maybe your final answer should be h + L/2.

Answer 3

Given:
distance travelled by ball = L
time taken by ball to cover this distance = t

initial velocity when ball was dropped u=0
let the total time taken by the ball from the instant when it was dropped till it was last seen by the observer be T.
when the person saw the ball at first instant, the velocity v1=0+g(T-t)
now,
L=(u1*t)+((g*t^2)/2)

also u1=g*(T-t)

so,
L=(g*(T-t))*t + ((g*t^2)/2) ------------ eq1

let the height upto which ball has fallen before dissapearing be h
so we have to calculate h

h= 0*T +((g*(T^2))/2)

=> h=(g*(T^2))/2 -------------- eq2

solving equation 1, we get

T=( (L - ( (g * (t^2) )/2 ) )/(g*t) ) + t
substituting this value of T in eq2, we get

h = ( ((2*L) + (g*(t^2)))^2 ) / (8*g*(t^2))

it is the total height from the top from when the ball was released upto the distance when tha balll was l;ast seen by the observer.......

Answer 4

some distance above you in meters or feet=velocity x time you saw it