Please help. College physics question!?

Question

This is due tonight and I DO NOT understand. How do you due this problem, and what is the answer?

"A freely falling object requires 1.90 seconds to fall the last 38.0 m before it reaches the ground. From what height did the object fall?"

This is not in my book. None of the formulas work because you have to know more variables than I am given. PLEASE help me.

Answer 1

A is the topmost point, velocity at this point is 0
B is the intermediate Position,velocity at this point is u1
C is the bottom position, velocity at this point is V

| A, u = 0
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| B, u1
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| C, V

Lets call the distance AB as S1
Lets call the distance BC as S

S = 38m

Step to find the velocity at position B

S = u1*t + (0.5)*g*t^2

38 = u1*1.9 + (0.5)*(9.8)*(1.9)*(1.9)

u1 = (38 - 17.69 )/1.9
u1 = 10.69 m/s

Step to find the distance AB

(u1)^2 = u^2 + 2*g*S1

(10.69)^2 = 2*9.8*S1 (Since u = 0)

S1 = 114.276/19.6
S1 = 5.83m
Total Height = AB + BC

= 5.83 + 38 = 43.38m

Answer 2

Ask an consultant interior the physics branch. diverse colleges have diverse names for various sequence of training they supply. working example, at my college, that they had concepts of Physics, general Physics, Classical Physics, and Honors Physics. I took general and that i'm rather helpful no different college teaches physics like that. So in straightforward terms somebody at your college can make it easier to be attentive to what the adaptation is. Random human beings on the internet have no theory.

Answer 3

Im stuck on some physics hw myself. The answer is 43.81. It took me a pretty long time, but im here waiting for someone to help me in physics, so w/e, right? So you use like 3 dvat equations. Then you solve for one variable at a time until you finally find ur answer. Ok, so, first, i used: x=x_0 +vt - 0.5at^2. I plugged in 1.9 for time, -9.81 for a, 0 for x, and 38 for x_0. i got the final velocity of 29.3195. You know that at the height of the drop, the initial velocity was 0, so i used that. So, next i used this equation: v^2=v_0^2 +2a(x-x_0). V=29.3195, x=0, a=-9.91, and v_0=0. Plug all those in, and then youll eventually get my answer: 43.81.

Answer 4

A) Calculate the average velocity for that last 1.90 seconds.
V = 38.0m / 1.90s
V = 20m/s

B) Add 0.95 (half of 1.90) seconds worth of acceleration to that average to get the velocity right when the object hit the ground (height of 0.0 m)
V final = 20m/s + (9.8m/s^2 * 0.95s)
V f = 20m/s + 9.31m/s
V f = 29.31m/s

C) Divide that final velocity by 9.8, to get the number of seconds it has been accelerating, and also in half to get the average velocity over the entire drop.
T total = (29.31m/s) / (9.8m/s^2)
T t = 2.99s

V avg = (29.31m/s) / 2
V a = 14.66m/s

D) Multiply the number of seconds it's been falling and the average velocity to get the original height.
H = 2.99s * 14.66m/s
H = 43.83m