Standard +0.3 This is a standard M2 projectiles question with routine application of the trajectory equation (part a is a bookwork 'show that'), energy conservation (straightforward substitution), and using the trajectory equation with known values. All techniques are standard textbook exercises requiring no novel insight, making it slightly easier than average.
A particle \(P\) is projected from a fixed point \(O\). The particle is projected with speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at angle \(\alpha\) above the horizontal. The particle moves freely under gravity. At the instant when the horizontal distance of \(P\) from \(O\) is \(x\) metres, \(P\) is \(y\) metres vertically above the level of \(O\).
Show that \(y = x \tan \alpha - \frac { g x ^ { 2 } } { 2 u ^ { 2 } } \left( 1 + \tan ^ { 2 } \alpha \right)\)
A small ball is projected from a fixed point \(A\) with speed \(U \mathrm {~ms} ^ { - 1 }\) at \(\theta ^ { \circ }\) above the horizontal.
The point \(B\) is on horizontal ground and is vertically below the point \(A\), with \(A B = 20 \mathrm {~m}\).
The ball hits the ground at the point \(C\), where \(B C = 30 \mathrm {~m}\), as shown in Figure 4.
\begin{figure}[h]
\end{figure}
The speed of the ball immediately before it hits the ground is \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)
The motion of the ball is modelled as that of a particle moving freely under gravity.
Use the principle of conservation of mechanical energy to find the value of \(U\).
\begin{enumerate}
\item A particle $P$ is projected from a fixed point $O$. The particle is projected with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at angle $\alpha$ above the horizontal. The particle moves freely under gravity. At the instant when the horizontal distance of $P$ from $O$ is $x$ metres, $P$ is $y$ metres vertically above the level of $O$.\\
(a) Show that $y = x \tan \alpha - \frac { g x ^ { 2 } } { 2 u ^ { 2 } } \left( 1 + \tan ^ { 2 } \alpha \right)$
\end{enumerate}
A small ball is projected from a fixed point $A$ with speed $U \mathrm {~ms} ^ { - 1 }$ at $\theta ^ { \circ }$ above the horizontal.\\
The point $B$ is on horizontal ground and is vertically below the point $A$, with $A B = 20 \mathrm {~m}$.\\
The ball hits the ground at the point $C$, where $B C = 30 \mathrm {~m}$, as shown in Figure 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ee5f81bc-1bdb-47a1-81e7-7e3cb8219e91-24_556_961_904_552}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
The speed of the ball immediately before it hits the ground is $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
The motion of the ball is modelled as that of a particle moving freely under gravity.\\
(b) Use the principle of conservation of mechanical energy to find the value of $U$.\\
(c) Find the value of $\theta$
\hfill \mbox{\textit{Edexcel M2 2023 Q8 [12]}}