Question

(113 A particle is projected from a tower of height 2 m
above ground with speed 3 m/s. Neglect any air
resistance to the particle's motion. Maximum
horizontal distance from the foot of the tower where
the particle can reach is (Take g = 10 m/s2)
(1) 1.7 m
(2) 1.9 m
0 (3) 2.1 m
(4) 2.3 m​

Answer 1

u=3m/s

g=-10m/s^2

v=0m/s

v^2=u^2+2as

0=9+2(-10) s

s=9/20m=0.45m

so, total height=2+0.45=2.45m

Answer 2

Answer:

2.1

Explanation:

-(3cos45)²+2(g)2=v²

40-9/2=v²

v=(71/2)½

v=u+at

(71/2)½-3/(2)½/10 =t

3cos30 ×t = d

3/(2)½(t)

Substituting the value of t we get

X= 2.14