 Hi and welcome to the session. Let us discuss the polyquestion. The question says a cricket ball is projected to the velocity of 29.4 m per second. Find first part is the greatest range on the horizontal plane and second part is the angle of projection to give a range of 44.10 m. In the question we have given that cricket ball is projected with the velocity of 29.4 m per second. That means initial velocity that is 0 is equal to 29.4 m per second. We have to find the greatest range on the horizontal plane. We know that greatest range which we denote by r is given by u squared by g. Now here u is equal to 29.4 m per second. So we have 29.4 whole squared by g is 9.8. This is equal to 29.4 into 29.4 divided by 9.8. On simplifying this we get 88.2. So greatest range on the horizontal plane is 88.2 meters. Now this is our answer to the first part. Second part we have to find the angle of projection to give a range of 44.10 m. Let alpha be the angle of projection for a range of 44.10 m. We know that range is given by u squared sine 2 alpha by g. Now here we are given r as 44.10 m. We have 44.10 equals to u is 29.4. We have 29.4 whole squared into sine 2 alpha by g is 9.8. Now this implies sine 2 alpha is equal to 44.10 into 9.8 divided by 29.4 into 29.4. On simplifying this we get 1 by 2. So we have sine 2 alpha equals to 1 by 2. This implies sine 2 alpha is equal to sine 30. This implies 2 alpha is equal to 30 degree and this implies alpha is equal to 15 degree. Hence our required answer of the first part is 88.2 meters and of the second part is 15 degree. So this completes the session. Bye and take care.