Moderate -0.3 This is a straightforward projectile motion problem where the vertical displacement equation is already provided, requiring only substitution to find time, then calculation of horizontal distance using basic kinematics. The problem-solving is routine: solve the given quadratic for t (taking the larger root for descending), find sin α from cos α, then multiply by horizontal velocity component. Slightly easier than average due to the given vertical equation and clear structure.
4 Sandy is throwing a stone at a plum tree. The stone is thrown from a point O at a speed of \(35 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(\alpha\) to the horizontal, where \(\cos \alpha = 0.96\). You are given that, \(t\) seconds after being thrown, the stone is \(\left( 9.8 t - 4.9 t ^ { 2 } \right) \mathrm { m }\) higher than O .
When descending, the stone hits a plum which is 3.675 m higher than O . Air resistance should be neglected.
Calculate the horizontal distance of the plum from O .
Equating given expression to their attempt at \(y\) to \(\pm 3.675\); if they attempt \(y\), allow sign errors, \(g=9.81\) etc, \(u=35\)
Solving \(4t^2 - 8t + 3 = 0\)
M1*
Dependent. Any method of solution of a 3-term quadratic
\(t = 0.5\) or \(t = 1.5\)
A1, F1
cao. Accept only larger root given; both roots shown and larger chosen provided both +ve. Dependent on 1st M1. [Award M1 M1 A1 for 1.5 seen WW]
Or method:
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
Complete method for total time
M1
Complete method for total time from motion in separate parts. Allow sign errors, \(g=9.81\) etc. Allow \(u=35\) initially only
\(0 = 35\times0.28 - 9.8t\), so \(t=1\)
A1
Time for 1st part
Time to drop is \(0.5\)
A1
Time for 2nd part
Total is \(1.5\) s
A1
cao
Then:
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
Horiz distance is \(35\times0.96t\)
B1
Use of \(x = u\cos\alpha t\). May be implied
Distance is \(35\times0.96\times1.5 = 50.4\) m
F1
FT their quoted \(t\) provided it is positive
6
# Question 4:
**Either method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3.675 = 9.8t - 4.9t^2$ | *M1 | Equating given expression to their attempt at $y$ to $\pm 3.675$; if they attempt $y$, allow sign errors, $g=9.81$ etc, $u=35$ |
| Solving $4t^2 - 8t + 3 = 0$ | M1* | Dependent. Any method of solution of a 3-term quadratic |
| $t = 0.5$ or $t = 1.5$ | A1, F1 | cao. Accept only larger root given; both roots shown and larger chosen provided both +ve. Dependent on 1st M1. [Award M1 M1 A1 for 1.5 seen WW] |
**Or method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method for total time | M1 | Complete method for total time from motion in separate parts. Allow sign errors, $g=9.81$ etc. Allow $u=35$ initially only |
| $0 = 35\times0.28 - 9.8t$, so $t=1$ | A1 | Time for 1st part |
| Time to drop is $0.5$ | A1 | Time for 2nd part |
| Total is $1.5$ s | A1 | cao |
**Then:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horiz distance is $35\times0.96t$ | B1 | Use of $x = u\cos\alpha t$. May be implied |
| Distance is $35\times0.96\times1.5 = 50.4$ m | F1 | FT their quoted $t$ provided it is positive |
| | **6** | |
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4 Sandy is throwing a stone at a plum tree. The stone is thrown from a point O at a speed of $35 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $\alpha$ to the horizontal, where $\cos \alpha = 0.96$. You are given that, $t$ seconds after being thrown, the stone is $\left( 9.8 t - 4.9 t ^ { 2 } \right) \mathrm { m }$ higher than O .
When descending, the stone hits a plum which is 3.675 m higher than O . Air resistance should be neglected.
Calculate the horizontal distance of the plum from O .
\hfill \mbox{\textit{OCR MEI M1 2009 Q4 [6]}}