1. The problem statement, all variables and given/known data A paintball gun launches a paintball off a cliff at an angle of elevation of 45°. The cliff is 165 m high. The paintball is initially moving at 180 m/s. Calculate the speed of the paintball as it hits the ground. I'm having trouble understanding how to solve this, I have posted my attempt. 2. Relevant equations Ek=1/2mv^2 Eg=mgh 3. The attempt at a solution I'm gonna find the kinetic energy using the initial velocity. So, cos 45 x 180 = 127.3 m/s. <---- This is velocity in the x direction Ek=1/2 m *(127.3)^2 =8102.6m Find height by subbing this into the Eg formula: 8102.6m=mgh 8102.6=9.8h 826.8=h So this is the height of the projectile, in reference to the cliff. The total height however should have the 165m of the cliff added: Therefore it is (826.8+165)=991.8 m This is the maximum height, I will use it to find the total energy at max height, so Eg=991.8gm =991.8(9.8)m =9719m Now I sub this into Ek to find the Velocity. Ek=9719m 1/2mv^2=9719m 1/2v^2=9719 v^2=19439 v=139 m/s The book answer says 190 m/s.