Standard +0.3 This is a standard projectile motion problem requiring resolution of velocity components and application of the constant acceleration formula. While it involves finding when speed equals a specific value 'for the second time' (requiring consideration of both ascending and descending phases), the method is straightforward: resolve initial velocity, use v² = u² + 2as to find the vertical component when speed is 25 m/s, then use v = u + at. It's slightly above average difficulty due to the multi-step nature and the need to identify the correct (second) solution, but remains a routine mechanics question.
A small ball \(B\) is projected with speed \(30\text{ m s}^{-1}\) at an angle of \(60°\) to the horizontal from a point on horizontal ground. Find the time after projection when the speed of \(B\) is \(25\text{ m s}^{-1}\) for the second time.
[4]
Question 1:
1 | v2 = 252 – (30cos60)2 | M1 | v = vertical velocity at the required point
v = (±) 20 | A1
–20 = 30sin60 – gt | M1 | Use v = u + at vertically
t = 4.6(0) s | A1
4
Question | Answer | Marks | Guidance
A small ball $B$ is projected with speed $30\text{ m s}^{-1}$ at an angle of $60°$ to the horizontal from a point on horizontal ground. Find the time after projection when the speed of $B$ is $25\text{ m s}^{-1}$ for the second time.
[4]
\hfill \mbox{\textit{CAIE M2 2018 Q1 [4]}}