Hello, I\'m having trouble with this physics problem A certain freely failing ob

Question

Hello, I'm having trouble with this physics problem

A certain freely failing object requires 2.00 s to travel the last 21.0 m before it hits the ground. From what heightabove the ground did it fall?

I used the y=Vot - 1/2gt2 equation to find theinital velocity and I got 21.0=Vo(2)-1/2(9.8)(2)2 and Igot Vo = -20.3 not sure if its suppose to be negative ornot. Than I used the V=Vo+at equation to find the finalvelocity and I got V=(-20.3)-(9.8)(2), V=-39.0. I than used theformula Y=Vavt to find the distance and substitued Vav forVo+V and here is the answer (-20.3)+(-39.9) andI got -30.1so
                                      2                                                             2
Y=-30.1(2.00) = 60.2m. Not sure if I'm going about itcorrectly. I could use some help.


Thank you!!!

Explanation / Answer

Distance travelled in last 2 seconds S = 21 m from the relation S = ut + ( 1/ 2) gt^ 2                       21 = u(2) + (4.9 * 2 ^ 2 ) from this initial velocity of the object just before 2 secondsof hitting   u = [ 21 - ( 4.9*2^2 ) ] / 2                                                                                                           = 0.7 m / s velocity of the object just befire hitting v = u+gt                                                             = 0.7 + ( 9.8 * 2 )                                                             = 20.3 m / s we know v = [ 2gH ] from this height H = v ^ 2 / 2g                             = 21.025 m

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---