A person lying 15 meters from a wall aims at a spot on the wall with an airgun. The spot is 7 meters high on the wall. The initial velocity of the bullet is 85 m/s. How far below the spot will the bullet strike if we disregard air resistance.

What I did first was to calculate the diagonal distance the bullet would travel using the Pythagorean theorem: sqrt(15^2 + 7^2) = 16.55 m. Then, to find the time it takes the bullet to strike the wall: t = 16.55 m / 85 m/s = 0.1947 s Then to find the angle of the bullet's trajectory: arctan(7/15) = 25 Then using the angle to find the initial vertical velocity: v_0y = 85 m/s * sin(25) = 35.92 m/s Then finally to calculate the height of the bullet when it strikes the wall: h = v_0y * t - at^2 / 2 = 35.92 m/s * 0.1947 s - 9.8 m/s^2 * (0.1947)^2 * 0.5 = 6.8 m

I get approximately 7 m - 6.8 m = 0.2 m, which is apparently wrong but I don't see my mistake.