AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Let u is the velocity when angle is 60∘ and v is the velocity when angle is 45∘ with the horizontal.
In projectile motion, the horizontal component of velocity remains constant throughout the motion. Therefore,
v‌cos‌45∘‌‌=u‌cos‌60∘
v×‌

1

√2

‌‌=140×‌

1

2

v‌‌=70√2m∕s
If t is the time after which projectile makes an angle of 45∘ then
v‌sin‌45∘‌‌=u‌sin‌60∘−gt
(70√2)‌

1

√2

‌‌=‌

√3

2

(140)−10t
10t‌‌=70√3−70
t‌‌=7(√3−1)
‌‌=5.124s