Concept:
h=2gu2sin2θ
Where h is the height.
Hmax=2gu2
Where Hmax is maximum height.
v = u + at
s=ut+21at2
s=ut−21at2
s=21(u+v)t
v2 = u2 + 2as
Where, v = final velocity, u= initial velocity, s = distance traveled by the body under motion,
a = acceleration of body under motion, and t = time taken by the body under motion.
Calculation:
Given: distance travelled by particle in 5th and 6th sec is equal in vertical motion of particle.
The distance travelled in 5th Sec:
h5=u−21g(2t−1)
⇒ h5=u−29g
The distance travelled in 6th sec:
h6=u−21g(2t−1)
⇒ h6=u−211g
So, h5 = -h6 (at t = 5 sec, particle will be at maximum height.)
⇒ u−29g=−u+211g
⇒ u = 5g
Maximum height attained,
Hmax=2gu2=2g(5g)2=225g=225×10=125;m
Hmax = 125 m
Alternative Method:
Time of flight = (t1 + t2 -1)
T = 5 + 6 – 1 = 10 sec
i.e it will take 5 sec to upward as well as downward journey.
In case of reverse journey particle initial velocity is zero. (Concept)
And time is 5 sec, acceleration is g = 10 m/s2.
s=ut+21at2
s=0×5+21×10×52
s = 125 m