Question 13
In this question use g=9.8g = 9.8g=9.8 m s−2^{-2}−2.
A ball is projected with speed U U\,U m s−1^{-1}−1 from a point OOO, 2 m above horizontal ground, at an angle α \alpha\,α above the horizontal. The ball is modelled as a particle and air resistance is modelled as negligible.
In its motion the ball reaches a maximum height of 3 m above the ground, and it passes through the point AAA, which is 0.8 m above the ground and a horizontal distance of 10 m from OOO.
a.[2]
Show that U2sin2α=19.6U^2\sin^2\alpha = 19.6U2sin2α=19.6.
b.[4]
By writing down expressions for the horizontal and vertical displacements of the ball at A A\,A in terms of t t\,t and eliminating ttt, show that 25tan2α−10tanα−1.2=025\tan^2\alpha - 10\tan\alpha - 1.2 = 025tan2α−10tanα−1.2=0.
c.[3]
Find the value of α \alpha\,α and the value of UUU.
3.3 Projectiles (A-level only) Questions
91 exam-style questions on OCR (MEI) A Level Maths 3.3 Projectiles (A-level only), covering 3.3.1 Model motion under gravity using vectors (A-level only), 3.3.2 Position, velocity, range and maximum height (A-level only), 3.3.3 Find the initial velocity of a projectile (A-level only), 3.3.4 Equation of the trajectory (A-level only), and 3.3.5 Solve simple projectile problems (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.