QCE Vault / Specialist Maths Vector calculus — Question 36 Original QCE Vault practice · 6 marks
Browse all questions Vectors revision Original practice exam Report an issue Q36 · Practice question Complex familiar 6 marks
QUESTION 36 (6 marks) A projectile is launched from level ground at 20 m s − 1 20\ \mathrm{m\,s^{-1}} 20 m s − 1 at 45 ∘ 45^\circ 4 5 ∘ above the horizontal. Take g = 10 m s − 2 g=10\ \mathrm{m\,s^{-2}} g = 10 m s − 2 and ignore air resistance. a) Use vector calculus to obtain the position vector r ( t ) \mathbf r(t) r ( t ) . [2 marks] b) Determine the maximum height.
[2 marks] c) Determine the horizontal range.
[2 marks] WORKED SOLUTION
Practice marking scheme 6 marks ANSWER (a) ( 10 2 t , 10 2 t − 5 t 2 ) (10\sqrt2\,t,10\sqrt2\,t-5t^2) ( 10 2 t , 10 2 t − 5 t 2 ) ; (b) 10 m 10\ \mathrm m 10 m ; (c) 40 m 40\ \mathrm m 40 m . Worked solution
(a) The acceleration is ( 0 , − 10 ) (0,-10) ( 0 , − 10 ) and initial velocity is ( 10 2 , 10 2 ) (10\sqrt2,10\sqrt2) ( 10 2 , 10 2 ) . Integrating, v ( t ) = ( 10 2 , 10 2 − 10 t ) . \mathbf v(t)=(10\sqrt2,10\sqrt2-10t). v ( t ) = ( 10 2 , 10 2 − 10 t ) . Integrating again with r ( 0 ) = ( 0 , 0 ) \mathbf r(0)=(0,0) r ( 0 ) = ( 0 , 0 ) gives r ( t ) = ( 10 2 t , 10 2 t − 5 t 2 ) . \mathbf r(t)=(10\sqrt2\,t,10\sqrt2\,t-5t^2). r ( t ) = ( 10 2 t , 10 2 t − 5 t 2 ) . (b) At maximum height, 10 2 − 10 t = 0 10\sqrt2-10t=0 10 2 − 10 t = 0 , so t = 2 t=\sqrt2 t = 2 . Substituting into the vertical position gives h = 20 − 10 = 10 m h=20-10=10\ \mathrm m h = 20 − 10 = 10 m . (c) The nonzero root of 10 2 t − 5 t 2 = 0 10\sqrt2\,t-5t^2=0 10 2 t − 5 t 2 = 0 is t = 2 2 t=2\sqrt2 t = 2 2 . Thus R = 10 2 ( 2 2 ) = 40 m R=10\sqrt2(2\sqrt2)=40\ \mathrm m R = 10 2 ( 2 2 ) = 40 m . Obtains the position vector using integration.
[2 marks] Determines the maximum-height time and height.
[2 marks] Determines flight time and range.
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