TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

The given situation of entire motion of ball is shown in the following figure. Initial speed of ball, u=20m∕s
Total time taken by the ball to reach at highest point P is t1, then from equation of vertical motion
v=u−gt⇒0=20−10t1⇒t1=2s
Maximum height attained by the ball is h, then from equation
h=ut−‌

1

2

gt2=20×2−‌

1

2

×10×22=40−20=20m
t2 be the time taken by the ball to reach at B from point P, then
h−5‌‌=0×t2+‌

1

2

gt22⇒20−5=0+‌

1

2

×10t22
15‌‌=5t22⇒t2=√3s
∴ Time taken by the ball during entire trip,
t=t1+t2=2+√3s