Question

4.
The velocity of a particle when it is at a maximum
height is
times of its velocity when at half of its
greatest height. The angle of projection is
(a) 300
(C) 450
(b) 600
(d) 900
5.
A particle is projected with a certain velocity. So
as to have the same horizontal range 'R'. If ty and
ty are the times taken to reach this point in the two
possible way.
(a) ty/t2 = 2R/g (b) tyt2 = 2R/g
(C) t+t2 =2R/g (d) ty- tg = 2R/g​

Answer 1

Explanation:

The velocity at the maximum height of projectile is half of its initial velocity projection. The angle of projection isWhen a projectile reaches maximum height, the vertical component of its velocity is momentarily zero (vy = 0 m/s). However, the horizontal component of its velocity is not zero.zero

At a projectile's highest point, its velocity is zero. At a projectile's highest point, its acceleration is zero

Answer 2

Answer:

answer of 4) 600

Explanation:

answer of 5) A) ty/t2 = 2R/g