Question 49
In this question use g=9.8g = 9.8g=9.8 m s−2^{-2}−2.
A ball, modelled as a particle, 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. 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.
3.2 Kinematics Questions
260 exam-style questions on OCR A Level Maths 3.2 Kinematics, covering 3.2.1 Language of kinematics, 3.2.2 Graphs in kinematics, 3.2.3 Displacement-time and velocity-time graphs, 3.2.4 Constant acceleration formulae, 3.2.5 Constant acceleration in two dimensions (A-level only), 3.2.6 Non-uniform acceleration in one dimension (A-level only), 3.2.7 Non-uniform acceleration in two dimensions (A-level only), 3.2.8 Motion under gravity using vectors (A-level only), and 3.2.9 Projectiles (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.