A particle is thrown up from a top of tower of 100m with the speed of 20m/s. g = 10m/s^2. Draw velocity-time graph and acceleration-time graph.
Answers
suppose sir it drops from a hight of x meters and which means
distance = 2 displacements so sir ,
=> H + (H-x) = 2 * x
=> 2H = 3x
=> x = 2/3 H
now sir , we get H
so sir now we will use v = u + at
and sir v is zero at maximum hight which means final velocity will become
zero at its maximum point
which means sir ,
v = u + at
0 = 20 - 10 *t g is negative sir because it is opposite to the motion
we get t = 2sec sir ,
now sir we will use ,
S= ut +1/2 at ^2 to find hight so sir
H = 20 ×2 - 1/2 ×10 × 4
which means sir ,
H = 40 - 20 which equals to = 20 sir
but sir x = 2/3 H so sir we put H here to find x
x= 2/3 ×20 = 40/3
Now sir , we just need to find out how long it takes for an object dropped
from
rest to fall a
and sir distance of H-x = 20 - 40/3 = 20/3
Using the equation S = ut + 1/2at^2, we will find the time for x
20/3 = 20×t - 1/2 10 ×t ^2
which means 5t^2 -20t + 20/3 =0
now i will multiply the equation by 3
15 t² - 60t +20 = 0
now sir i will divide the equation by 5
3 t²-12t + 4 = 0
now sir this equation is quadratic in t we will find its root by quadratic
formula
t = [-b +,- √b²-4ac] / 2a and here a = 3 b = -12 and c = 4
t= [ 12 +,- √ 12²-4×4×3 ] /2×3
t= [ 12 + ,-√144-48]/6 which means
t= [ 12 +,-√96]/6
√96 = √16 ×6= 4√6
now sir , we get t = [12 +,-√ 4√6/6 ]/6
which means we get t = Thus, 2 seconds to get to max height and 2/√3
secs to fall, giving a total time requirement of 2 + 2/√3
after solving the equation
HOPE THIS HELPS YOU SIR !
# regards Brainly helper
here see that acceleration is constant and in always in downward direction so it is taken as negative...
by convention it is supposed that upward direction is positive and downward as negative...
see attachment....
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hope it will help u
