1. The problem statement, all variables and given/known data A bullet is fired from a gun fixed at a point O with speed v m/s at an angle Θ to the horizontal, At the instant of firing , a moving target is 10m vertically above O and travelling with constant speed 42√ 2 m/s at a constant angle 45 degrees to the horizontal.The bullet and target move in the same plane. (i) if v = 70 m/s, show that if the bullet is to hit the target then tanΘ = 4/3 (ii) Find at what time after firing does the bullet strike the target and calculate the horizontal distance of the bullet from O 2. Relevant equations s=ut + (1/2)gt^2 3. The attempt at a solution If they are to collide then there horizontal velocities must be equal. 70cosΘ = 42√2(cos45) ............. cosΘ = 42/70 =3/5 using Pythagoras' sinΘ = 4/5 and tanΘ = 4/3 (ii) There height must be the same when they collide h1 = h2 42√ 2(sin45)t - (1/2)gt^2 + 10 = 70(4/5)t - (1/2)gt^2 (1/2)gt^2 cancel 42t + 10 = 56t ...................... t = 10/14 seconds horizontal distance of the bullet from O 70(3/5)(10/14) = 30m My book says the answers for (ii) are 10/7 seconds and 60m. Can anybody spot if I went wrong at any point? Any help would be appreciated.