Pre-U Pre-U 9794/3 2017 June — Question 7 9 marks

Exam BoardPre-U
ModulePre-U 9794/3 (Pre-U Mathematics Paper 3)
Year2017
SessionJune
Marks9
TopicProjectiles
TypeHorizontal projection from height
DifficultyModerate -0.8 This is a straightforward two-part projectiles question using standard SUVAT equations with horizontal projection. Part (i) requires finding time from vertical motion then using horizontal distance; part (ii) needs resolving final velocity components. Both are routine textbook exercises requiring only direct application of memorized methods with no problem-solving insight or geometric complexity.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model

7 A building 33.8 m high stands on horizontal ground. A particle is projected horizontally from the top of the building and hits the ground 31.2 m away.
  1. Find the initial speed of the particle.
  2. Find the magnitude and direction of the velocity of the particle when it hits the ground.

Question 7(i)
Alternative version 1:
- Vertical: \(-5t^2 = -33.8\) B1 (Allow absence of both minus signs.)
- \(\therefore t = \sqrt{\dfrac{33.8}{5}} = 2.6\) (sec) B1 (cao)
- Horizontal: \(2.6u = 31.2\) M1
- \(\therefore u = 12\) (ms\(^{-1}\)) A1 (FT *their* \(t\).)
Alternative version 2:
- Horizontal: \(ut = 31.2\) B1
- \(33.8 = 5 \times \dfrac{31.2^2}{u^2}\) M1 (Eliminate \(t\).)
- \(\therefore u = \sqrt{\dfrac{5 \times 31.2^2}{33.8}} = 12\) (ms\(^{-1}\)) A1
Question 7(ii)
- \(v_x = 12\)
- Either \(v_y = (0) - 10 \times 2.6 = -26\) B1 (FT *their* \(t\). Allow absence of minus sign.)
- Or \(v_y = \sqrt{(0^2) + 2 \times 10 \times 33.8} = (-)26\)
AnswerMarks Guidance
- \(\thereforev = \sqrt{12^2 + (-26)^2} = \sqrt{820} = 28.635\ldots\) M1
- \(\approx 28.6\) (ms\(^{-1}\)) A1 (FT *their* \(v_y\) and/or \(u\).)
- \(\theta = \tan^{-1}\left(\dfrac{-26}{12}\right)\) M1
- \(= -65.2°\) A1 (Must be negative or have reference to the horizontal, e.g. "below …". FT *their* \(v_y\) and/or \(u\).)
Total: 9 marks
**Question 7(i)**

**Alternative version 1:**
- Vertical: $-5t^2 = -33.8$ **B1** (Allow absence of both minus signs.)
- $\therefore t = \sqrt{\dfrac{33.8}{5}} = 2.6$ (sec) **B1** (cao)
- Horizontal: $2.6u = 31.2$ **M1**
- $\therefore u = 12$ (ms$^{-1}$) **A1** (FT *their* $t$.)

**Alternative version 2:**
- Horizontal: $ut = 31.2$ **B1**
- $33.8 = 5 \times \dfrac{31.2^2}{u^2}$ **M1** (Eliminate $t$.)
- $\therefore u = \sqrt{\dfrac{5 \times 31.2^2}{33.8}} = 12$ (ms$^{-1}$) **A1**

**Question 7(ii)**
- $v_x = 12$
- Either $v_y = (0) - 10 \times 2.6 = -26$ **B1** (FT *their* $t$. Allow absence of minus sign.)
- Or $v_y = \sqrt{(0^2) + 2 \times 10 \times 33.8} = (-)26$
- $\therefore |v| = \sqrt{12^2 + (-26)^2} = \sqrt{820} = 28.635\ldots$ **M1**
- $\approx 28.6$ (ms$^{-1}$) **A1** (FT *their* $v_y$ and/or $u$.)
- $\theta = \tan^{-1}\left(\dfrac{-26}{12}\right)$ **M1**
- $= -65.2°$ **A1** (Must be negative or have reference to the horizontal, e.g. "below …". FT *their* $v_y$ and/or $u$.)

**Total: 9 marks**
7 A building 33.8 m high stands on horizontal ground. A particle is projected horizontally from the top of the building and hits the ground 31.2 m away.\\
(i) Find the initial speed of the particle.\\
(ii) Find the magnitude and direction of the velocity of the particle when it hits the ground.

\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2017 Q7 [9]}}