An object is dropped from a height H above the ground. This free-falling object requires 0.904 s to travel the last 34 m before hitting the ground.

From what height H above the ground did
the object fall? The acceleration of gravity is 9.8 m/s^2.
Answer in units of m

i understand that t= .904s
distance is 34m
and a = 9.8m/s^2
But I'm not sure what else to do here?

H=g•t²/2

H-34 = g• (t-0.904)²/2

g•t²/2 – 34 = 0,5•g•(t²-2•0.904•t +0.904²)
8.86t=38
t=38.8.86 = 4.3 s.
H= g•t²/2 = 9.8•4.3²/2 = 90 m

To calculate the initial height (H) from which the object fell, we can use the equations of motion for an object in free-fall.

First, we need to calculate the final velocity (Vf) of the object when it hits the ground. We know that the initial velocity (Vi) is 0 because the object is dropped. The equation to calculate the final velocity is:

Vf = Vi + at

Since Vi = 0 for a dropped object, we simplify the equation to:

Vf = at

Plugging in the given values, we have:

Vf = 9.8 m/s^2 * 0.904 s
Vf = 8.8736 m/s

Next, we need to calculate the time (t) taken by the object to fall from height H. We know that the distance traveled (d) during this time is 34 m. The equation to calculate the distance traveled is:

d = (Vi * t) + (0.5 * a * t^2)

Again, since Vi = 0, we simplify the equation to:

d = 0.5 * a * t^2

Plugging in the given values, we have:

34 m = 0.5 * 9.8 m/s^2 * t^2

Rearranging the equation gives us:

t^2 = (34 m * 2) / 9.8 m/s^2
t^2 = 6.9388 s^2

To solve for t, we take the square root of both sides:

t = √(6.9388 s^2)
t ≈ 2.633 s

Now that we know the time taken to fall from height H is 2.633 s, we can use this value to calculate H. The equation to calculate H is:

H = Vi * t + 0.5 * a * t^2

Since Vi = 0, we simplify the equation to:

H = 0.5 * a * t^2

Plugging in the given values, we have:

H = 0.5 * 9.8 m/s^2 * (2.633 s)^2
H ≈ 33.07 m

Therefore, the object fell from a height of approximately 33.07 m above the ground.