# 2 questions on Motion along a straight line

Question on "Motion along a straight line"

[edit: finally got the first one already so I erased it ...]

Hey guys, I really appreciate this forum and all that are taking the time to help out. I wish to be part of this community and help out others in the future as well. I had a question on Motion along a straight line.

An object is dropped from a height H above the ground. This free falling object requires 0.952 s to travel the last 46m before hitting the ground. The acceleration of grav. is 9.8 m/s^2. From what height H above the ground did the object fall. (answer in m).

I tried this a couple of times by getting the velocity using d = rt and setting vf = 0 and g = 9.8 and then using the d = vit + 1/2at^2 but again, prob. not the right approach. I think I'm just losing my head with all this physics because I got the harder problems right in this set of problems. If someone could help me on this, I would really appreciate it. I've been trying to get it for an hour now but am still unable to. The thing is, I have to input these answers online, so even if I'm off by a little, I don't know and I get points taken off for each wrong attempt, so I'm trying my best to get them right the first time. Well, thanks in advance guys. Hope to hear from you all.

Any suggestions would be appreciated; I just need some help on as to how to tackle this problem ...

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Fermat
Homework Helper
You were almost there. Did you use the eqn "d = vit + 1/2at^2" ? where vi is the velocity 46m above the ground and d = 46m and t = 0.952s ?
This woud give vi = 43.65 m/s.

How far does a body need to fall, from rest, under gravity, to reach a velocity of 46.53 m/s ?
Add this height onto the 46m and you're done.

Fermat,

Thanks a million bud. You just made that problem look insanely easy after I worked on it for so long. I guess my mind just doesnt function the way yours does but I guess I just have to look more closely at the information I have. Thanks again, I got the answer. It's approximately 143.23m.