Question

Dued Monday please help.

1. An object falls from rest on a high tower and takes 5.0 s to hit the ground. Calculate the object’s position from the top of the tower at 1.0 s intervals. Make a position-time graph for the object’s motion. In your response, show what you are given, the equation that you used, any algebra required, a table of data, and your graph.
g = 9.8 m/s2

Answer 1

@rajat97 can you help?

Answer 2

yup i'm here

Answer 3

okay , here's the answer to your question
if you observe the question carefully , you'll get that the object does vertical rectilinear and uniformly accelerated motion.
it means that the motion performed by the object is a vertical straight line and the acceleration is constant i.e. the acceleration due to gravity(you may know this)
so to deal with the first question , you should use the kinematical equation
s=ut+1/2 at^2
where u=initial velocity=0(here)
t=time
a=acceleration = acceleration due to gravity(here)=-9.8m/s^2(-ve due to downard direction)
for the first second,
s=0x1-1/2 x 9.8 x 1x1=-4.8m(s=-ve as it is displacement i.e. a vector quantity if it is asked to find the distance , then the distance in the first second will be 4.8m and not -4.8 m as distance cannot be -ve(i'll talk about it later))
so you can find it for the rest of the 4 seconds
for the next part of the question(show what you are given)
you can write about the specified and the hidden things
specified things are the time and nothing else
and hidden things are the initial velocity=0 (object released from rest)
acceleration=-9.8m/s^2(object dropped from top of a tower so it is well understood that it is on the earth)
so here the second part ends
for the third part, you can specify the kinematical equation s=ut+1/2 at^2
and for the algebra used, you can write about the calculations and the squaring (the equation has t^2)
and i dont think that this part has anything more than this
next, for the next part, i think you have to prepare a table of time and position of the object(idk much about it)
and for the last part, the figure is a parabola as for a unifomly accelerated motion , a graph of position v/s time is always parabola. Or else , you can also see the equation s=ut+1/2 at^2
if you observe, you will come to know that this is a quadratic equation in t
so a graph for a quadratic equation is a parabola. the first one is enough for your question
and i've attached the graph in this post
look,
you are said to make a table of the data that you get
you get the position of the particle at different time by solving the first part so you are expected to make a table of that data i.e. you table should depict the position of the object from the top of the tower at different time
and you just need to make it down onto you graph
and i want you to do it on your own
hope this helps you :)