 Hi and welcome to the session. Let's work out the following question. The question says the particle is projected so as to raise the tops of two walls each of height 10 meter at distances for 15 meter and 45 meter respectively from the point of projection. Prove that the angle of projection is standard verse 8 by 9. Now this is how we interpret the question. It's given that a particle is projected so as to raise the tops of two walls. Now each of them is 10 meter high at distances 15 meter and 45 meter from the point of projection that is O. So let us see the solution to this question. First of all let U and alpha be the velocity and angle of projection then the equation of trajectory will be y equals to x tan alpha minus gx square divided by 2u square cos square alpha and this we call equation 1. Now according to the question the particle is passing through the point B that is 1510 that is this is the point B and coordinates of point D are 4510 if O is considered as 00 so it must satisfy the equation of the projectile that is equation 1. Therefore 10 should be equal to 15 tan alpha minus g into 15 square divided by 2u square cos square alpha and this we call equation 2. So also we have 10 equals to 45 tan alpha minus g into 45 square divided by 2u square cos square alpha this we call equation 3. Now multiplying equation 2 by 9 and subtract 3 from it we get 90 minus 10 will be equal to 15 into 9 minus 45 tan alpha minus g into 2025 minus 2025 divided by 2u square cos square alpha. Now this gets cancelled so this term becomes 0 and this implies 80 is equal to 90 tan alpha and this implies that tan alpha is equal to 80 divided by 90 that is further equal to 8 by 9. So this implies that alpha is equal to tan inverse 8 by 9. Now this is the angle of projection and this is what we were supposed to prove in this question. I hope that you understood the solution and enjoyed the session. Have a good day.