TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

The given situation is as shown in figure,Let the particle reaches point P in time t. Then, horizontal distance,
x=u‌cos‌θ‌t⇒10=u‌cos‌45∘×t
⇒‌‌t=‌

10√2

u

...(i)
and vertical distance,
y=u‌sin‌θ‌t−‌

1

2

gt2
Substituting value of t from Eq. (i) into Eq. (ii), we get
⇒‌‌7.5=10−‌

1000

u2

⇒‌‌u2=‌

1000

2.5

=400 or u=20m/s
Hence, the initial speed of particle is 20m/s.