Question 1: Projectile Motion – Particle Projected at an Angle (Practice Question)
(a)
Find the time it takes the particle to reach the greatest height above the ground.
(2 Marks) 
(b)
Hence determine the distance $OP$.
The particle reaches a height $H$ meters above ground, $1$ second after leaving $O$.
(2 Marks) 
(c)
Find the value of $H$.
(2 Marks) 
(d)
Hence calculate the length of time for which the height of the particle above the ground is greater than $H$.
(2 Marks) Divide the Motion into Two directions "x" and "y". Since the particle lands at the same height, the displacement in "x" direction is zero.
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Correct Result
a) Time to greatest height $$t = \frac{12}{7} \text{ s} \quad (\approx 1.71 \text{ s})$$ b) Horizontal distance $OP$ $$|OP| = 76.8 \text{ m}$$ c) Height $H$ at $t=1$ $$H = 11.9 \text{ m}$$ d) Duration for which height $> H$ $$\Delta t = \frac{10}{7} \text{ s} \quad (\approx 1.43 \text{ s})$$
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Step-by-Step
Solution for Part A
### **Part (a): Time to Reach Greatest Height**
**Method:** Analyze the vertical motion using the equation $v = u + at$.
* **Initial vertical velocity ($u_y$):** $28 \sin \alpha = 28 \times \frac{3}{5} = 16.8 \text{ m s}^{-1}$
Solution for Part B
### **Part (a): Time to Reach Greatest Height**
**Method:** Analyze the vertical motion using the equation $v = u + at$.
* **Initial vertical velocity ($u_y$):** $28 \sin \alpha = 28 \times \frac{3}{5} = 16.8 \text{ m s}^{-1}$
Solution for Part C
### **Part (a): Time to Reach Greatest Height**
**Method:** Analyze the vertical motion using the equation $v = u + at$.
* **Initial vertical velocity ($u_y$):** $28 \sin \alpha = 28 \times \frac{3}{5} = 16.8 \text{ m s}^{-1}$
Solution for Part D
### **Part (a): Time to Reach Greatest Height**
**Method:** Analyze the vertical motion using the equation $v = u + at$.
* **Initial vertical velocity ($u_y$):** $28 \sin \alpha = 28 \times \frac{3}{5} = 16.8 \text{ m s}^{-1}$
Full Solution
### **Part (a): Time to Reach Greatest Height**
**Method:** Analyze the vertical motion using the equation $v = u + at$.
* **Initial vertical velocity ($u_y$):** $28 \sin \alpha = 28 \times \frac{3}{5} = 16.8 \text{ m s}^{-1}$
* **Vertical acceleration ($a$):** $-9.8 \text{ m s}^{-2}$
* **Final vertical velocity ($v$):** $0 \text{ m s}^{-1}$ (at max height)
$$
\begin{aligned}
v &= u + at \\
0 &= 16.8 - 9.8t \\
9.8t &= 16.8 \\
t &= \frac{16.8}{9.8} \\
\end{aligned}
$$
$$
\boxed{t = \frac{12}{7} \text{ s} \quad (\approx 1.71 \text{ s})}
$$
***
### **Part (b): Horizontal Distance OP**
**Method:** Use the symmetry of projectile motion to find the total flight time, then calculate horizontal distance.
**1. Find Total Flight Time**
By symmetry, the total time of flight is twice the time to reach the greatest height.
$$
\text{Total Time} = 2 \times \frac{12}{7} = \frac{24}{7} \text{ s}
$$
**2. Calculate Distance**
Using $\text{Distance} = \text{Speed} \times \text{Time}$ for the horizontal motion (where acceleration is zero).
* **Horizontal velocity ($u_x$):** $28 \cos \alpha = 28 \times \frac{4}{5} = 22.4 \text{ m s}^{-1}$
$$
\begin{aligned}
|OP| &= 22.4 \times \frac{24}{7} \\
\end{aligned}
$$
$$
\boxed{|OP| = 76.8 \text{ m}}
$$
***
### **Part (c): Value of H**
**Method:** Analyze the vertical motion again to find displacement ($s$) at $t=1$.
* **Variables:** $u = 16.8$, $a = -9.8$, $t = 1$.
* **Equation:** $s = ut + \frac{1}{2}at^2$
$$
\begin{aligned}
s &= 16.8(1) + \frac{1}{2}(-9.8)(1)^2 \\
s &= 16.8 - 4.9 \\
\end{aligned}
$$
$$
\boxed{H = 11.9 \text{ m}}
$$
***
### **Part (d): Duration Height > H**
**Method:** The particle takes $1$ second to reach height $H$ on the way up. By symmetry, it will take another $1$ second to fall from height $H$ back to the ground. We subtract these "tails" from the total flight time.
$$
\begin{aligned}
\text{Required Time} &= \text{Total Flight Time} - (2 \times 1) \\
&= \frac{24}{7} - 2 \\
&= \frac{24}{7} - \frac{14}{7} \\
\end{aligned}
$$
$$
\boxed{\Delta t = \frac{10}{7} \text{ s} \quad (\approx 1.43 \text{ s})}
$$