Question

A particle is projected from ground level with an initial velocity of 35m/s at an angle of tan^-1 3/4 to the horizontal find the time for which the particle is more than 2m above the ground

Answer 1

Ans. is 4 sec.... solution is in the pic :)

Answer 2

A particle is projected from ground level with an initial velocity 35 m/s at an angle of tan^-1(3/4) with horizontal.
so, vertical component of velocity = 35sin(tan^-1(3/4)) = 35 × 3/5 = 21 m/s
horizontal component of velocity = 35cos(tan^-1(3/4)) = 35 × 4/5 = 28 m/s

now, use formula , $Y=u_yt+\frac{1}{2}a_yt^2$ in vertical direction.

here, y = 2 m , $u_y=21m/s$
t = ? and $a_y=-g$
so, 2 = 21 × t - 1/2 × 10 × t²
2 = 21t - 5t²
5t² - 21t + 2 = 0
t = {21 ± √(21² - 2(5)(2))}/10
t = (21 ± √421)/10
t ≈ 0.097 , 4.1025

hence, there are two value of t , because in ascending motion, first time particle reaches 2m from ground and then descending motion 2nd time appeared at 2m above from the ground .