a ball is thrown vertically upwards from ground with an initial velocity of 48 feet per second. the position function is given by s(t)= -16t2+48t (a) after how many seconds does the ball attains its maximum height? (b) what is the maximum height the ball attains? (c) after how many second does the ball return to the ground?
Wrong Section.
no its a math problem just cuz its physics related doesnt mean that its in the wrong section
max height = 117.55 feet
using physics or math.
phy
that wont work i think they dont actually assume 9.81
if its not given we have to assume g=9.81
wht level of ques is this?? A-levls?
please show all work
v^2=u^2+2ah a=-g v=0 therefore h=u^2/2g
kis level ka question hai yeh
If that works go for it if not then its gonna be a tough one. What you do is you find the anti-derivative of the position function to find the distance traveled but you use the initial equation to find the amount of time before it stops moving. so when the position function=0 then the ball has reached the top and that is the T for part (a). By that same measure the time it takes to come back down is the time it took to get to that point. so the T you found at that point should be the half the value of T for part (c). Using the T from part (a) plug it into the anti-derivative of the position function and you should get the height.
quadratic function
we can acceleration by differntianing it twice with respect to 't'
no just once.
if we do it once we will get velocity in terms of t
then im making it way to complicated.
ans for C. is 9.79 sec.
C
ball will atain max height in 4.89 sec.
this is what i got. the time is 1.5 to reach the top and 3 secs to reach the ground again. plug 1.5 to find the height. 48*t=16*t^2 when t = 3 so that is when the position is zero again.
thats why you need the quadratic to solve for when it would equal zero.
what do u think halimabegum??????
I am confused!
wats ur point of view of solving it....?
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