hi..need help physics problem...?

Question

the time a male bungee jumper is freely falling is 1.5 seconds.
1. what is the velocity of the jumper at the end of 1.5seconds?
2. how far does he fall?




please with solutions....i cant do it..please thnks

Answer 1

1. a=(v-u)/t
a or g=10
u=0
t=1.5
v=?

so v=at+u
v=(10)(1.5)+0
v=15

2. v=s/t
s=vt
=(15)(1.5)
=22.5m
...hope u get it

Answer 2

Wow! If I had an actual *policy*, I'd be dangerous! Seriously, though, I've kind of wandered around a bit in policy-space, over the year+ that I've been on here. I look for questions that interest me, and that I feel I can make a contribution toward the asker's understanding of. I am especially attracted to questions that add, "... don't just give me the answer, I want to know how to do it." That's pure gold! But even if the question seems to be just looking for a quick answer, not caring about methods, I will still try, if I answer, to impart something of the method, or even generalize the solution, and add, "with this technique you should be able to handle any similar problem you encounter." Perhaps in the hope that at least a few of these askers will realize that a grasp of the right techniques can make their lives easier than constantly posting the same sorts of questions. And if the asker comes back with, 'don't understand that completely, could you give more detail?', I make an effort to transmit some understanding. I welcome such dialogs when they can actually have an impact. In the case of multiple problems in a single Y!A question, I guess like many of you, that has a tendency to put me off. But I find that when a question has unusually many parts, they are usually each more like morsels or tidbits, so sometimes I'll go through answering them all, or most of them, and remark, "Are you seeing a pattern here?" Finally, I almost always give a TD to an answer that just says, "Do your own @#$% homework." Anyone who is going to take that attitude, shouldn't be posting at all. BTW, MD, are you going to share your policy with us, perhaps at the close of the answer period? What are your own thoughts on this?

Answer 3

If the fall is purely due to gravity with no other force acting on him, then the equation v = u + g*t holds, where u = initial velocity, v = end velocity, g = acceleration due to gravity, and t the time taken.
In your particular instance, u = 0, g = 9.8 m/sec^2, t = 1.5 sec, giving v = 9.8 * 1.5 = 14.715 m/sec
The second question can be answered by using the equation
v^2 = u^2 + 2* g* h, where h = distance travelled.
From the above, it follows that v^2 = 2* 9.81 * h,
or h = 14.715 * 14.715/(2*9.81) = 11.04 m
[N.B. Check the calculations yourself, as my calculator is not behaving too well this morning]

Answer 4

The acceleration of gravity is 32 feet per second per second. So the velocity after 1.5 seconds (ignoring wind resistance) is 1.5 times 32, or 48 feet per second.
The distance fallen is given by (1/2) (g)(t squared) where g is the acceleration of gravity and t is the time. That's (1/2)(32)((1.5)squared) or, 36 feet.
Caution!! Please don't try this on your own.

Answer 5

assuming Acceleration due to gravity i.e. g = 9,8 metre/second square.
velocity = g*t
velocity at 1.5 seconds = 9.8*1.5 + 147

Distance travelled = (1/2)*g*t^2
= 11.025 meters

Answer 6

v=at ,a=g=10m/s2
v(1)=10*1.5=15m/s
v(1)^2=2gd
d=v(1)^2/2g
d=225/20
d=11.25m

Answer 7

please search in 8'th class book.
sorry.