To solve this problem, we analyze the motion of both the projectile and the particle \( P \) at the highest point \( H \).
### **Step 1: Understanding Projectile Motion**
A projectile follows a parabolic trajectory under gravity. The key properties at the highest point \( H \) are:
- **Velocity at \( H \)**: The horizontal component of velocity remains unchanged, while the vertical component becomes zero.
- **Acceleration at \( H \)**: The acceleration due to gravity acts downward.
Given:
- Initial speed \( u = 20 \) m/s
- Launch angle \( \theta = 60^\circ \)
- Gravitational acceleration \( g = 10 \) m/s²
The initial velocity components:
- \( u_x = u \cos 60^\circ = 20 \times \frac{1}{2} = 10 \) m/s
- \( u_y = u \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \) m/s
At the highest point \( H \), \( v_y = 0 \), so the velocity of the projectile is purely horizontal:
\( v_H = 10 \) m/s (along the x-axis).
### **Step 2: Understanding Particle \( P \) at \( H \)**
Given:
- Speed of \( P \) at \( H \): \( v_P = 20 \) m/s
- Acceleration magnitude \( a_P = 50 \) m/s²
Since \( P \) moves along the same trajectory as the projectile but with a varying speed, its velocity vector at \( H \) must be purely horizontal.
So, velocity vector:
\[ \mathbf{v_P} = (20 \hat{i}) \text{ m/s} \]
(where \( \hat{i} \) is the unit vector along the horizontal direction).
### **Step 3: Finding Acceleration Vector \( a_P \)**
Acceleration consists of two components:
1. **Tangential Acceleration \( a_T \)**: Responsible for changing the magnitude of speed.
2. **Normal Acceleration \( a_N \)**: Responsible for changing direction.
Since \( P \) follows the same trajectory as the projectile, its normal acceleration must be purely due to gravity in the downward direction.
Thus:
- Normal acceleration \( a_N = g = 10 \) m/s² (downward, along \( \hat{j} \)).
- Total acceleration magnitude \( a_P = 50 \) m/s².
Using Pythagoras theorem to find tangential acceleration \( a_T \):
\[
a_P^2 = a_T^2 + a_N^2
\]
\[
50^2 = a_T^2 + 10^2
\]
\[
2500 = a_T^2 + 100
\]
\[
a_T^2 = 2400
\]
\[
a_T = \sqrt{2400} = 20\sqrt{6} \text{ m/s²}
\]
Since the speed of \( P \) is increasing, \( a_T \) must be in the direction of velocity, which is along the x-axis.
Thus, the acceleration vector at \( H \) is:
\[
\mathbf{a_P} = (20\sqrt{6} \hat{i} - 10 \hat{j}) \text{ m/s²}
\]
