BSc Nursing GNM 2020 Solved Paper

Given, time of flight,
t=5s
Horizontal range, R=200m
We know that, the time period of a projectile
t=‌

2v‌sin‌θ

g

⇒‌‌v=‌

tg

2‌sin‌θ

. . . (i)
and the range of projectile
R=‌

v2‌sin‌2‌θ

g

. . . (ii)
Here, from Eqs. (i) and (ii), we get
‌

t2g2×2‌sin‌θ‌cos‌θ

4sin2θ×g

‌‌=‌

25×10

2

‌cot‌θ
⇒‌‌‌‌tan‌θ‌‌=‌

5

8

⇒‌‌θ‌‌=tan−1(‌

5

8

)