A projectile is launched at ground level with an initial speed of 53.0 m/s at an angle of 35.0° above the horizontal. It strikes a target above the ground 2.50 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

x = 108.5 m      y = 45.37 m,  

Explanation:

This is a projectile launching exercise, the horizontal distance it reaches is called the range and can be found by the expression

           x = v₀ₓ t

the initial hairiness is

          v₀ₓ = v₀ cos θ

           x = v₀ cos θ  t

  let's calculate

           x = 53 cos 35 2.50

           x = 108.5 m

the height of the projectile can be calculated with

            y =v₀ t - ½ g t²

             y = vo sin tea t - ½ g t²

            y = 53 sin 35 2.50 - ½ 9.8 2.5²

            y = 45.37 m

nah.   i'm just sleepy.