Question

how to prove it that 3rd particle takes √ t1.t2​

Answer 1

here is your answer

= V0 cos(θ) t       y = V0 sin(θ) t - (1/2) g t2

In the problem V0 = 20 m/s, θ = 25° and g = 9.8 m/s2.  

The height of the projectile is given by the component y, and it reaches its maximum value when the component Vy is equal to zero. That is when the projectile changes from moving upward to moving downward.(see figure above) and also the animation of the projectile.  

Vy = V0 sin(θ) - g t = 0  

solve for t  

t = V0 sin(θ) / g = 20 sin(25°) / 9.8 = 0.86 seconds  

Find the maximum height by substituting t by 0.86 seconds in the formula for y  

maximum height y (0.86) = 20 sin(25°)(0.86) - (1/2) (9.8) (0.86) 2 = 3.64 meters

hope this will help u:)