To solve the problem step by step, we will analyze the projectile motion of the stone and use the given information to find the velocity at the highest point.
### Step 1: Understand the Problem
We know that the stone is projected at an angle of \(53^\circ\) with the horizontal, and we need to find its velocity at the highest point of its trajectory. We are given that 2 seconds before reaching the maximum height, the stone's velocity makes an angle of \(53^\circ\) with the horizontal.
### Step 2: Determine the Time to Reach Maximum Height
Since the stone reaches its maximum height 2 seconds after the given time, we can conclude that the total time taken to reach the maximum height is \(t = 2\) seconds. Therefore, the time to reach the maximum height from the point where the angle is \(53^\circ\) is \(2\) seconds.
### Step 3: Use the Vertical Motion Equation
At the maximum height, the vertical component of the velocity becomes zero. We can use the following kinematic equation for vertical motion:
\[
v = u + at
\]
Where:
- \(v\) is the final vertical velocity (0 at maximum height),
- \(u\) is the initial vertical velocity,
- \(a\) is the acceleration due to gravity (which is \(-g\)),
- \(t\) is the time taken to reach the maximum height.
Rearranging gives:
\[
0 = u - gt
\]
Thus,
\[
u = gt
\]
### Step 4: Find the Initial Vertical Velocity
Given that \(t = 2\) seconds and \(g \approx 10 \, \text{m/s}^2\):
\[
u = 10 \times 2 = 20 \, \text{m/s}
\]
### Step 5: Relate Vertical Velocity to Initial Velocity
The vertical component of the initial velocity can be expressed as:
\[
u_y = u \sin(53^\circ)
\]
We know that:
\[
u_y = 20 \, \text{m/s}
\]
### Step 6: Calculate the Initial Velocity \(u\)
Using \(\sin(53^\circ) = \frac{4}{5}\):
\[
u \sin(53^\circ) = 20
\]
Substituting the value of \(\sin(53^\circ)\):
\[
u \cdot \frac{4}{5} = 20
\]
Solving for \(u\):
\[
u = \frac{20 \times 5}{4} = 25 \, \text{m/s}
\]
### Step 7: Find the Horizontal Component of Velocity
The horizontal component of the initial velocity is given by:
\[
u_x = u \cos(53^\circ)
\]
Using \(\cos(53^\circ) = \frac{3}{5}\):
\[
u_x = 25 \cdot \frac{3}{5} = 15 \, \text{m/s}
\]
### Step 8: Determine the Velocity at the Highest Point
At the highest point, the vertical component of the velocity is \(0\), and the horizontal component remains unchanged. Therefore, the velocity at the highest point is:
\[
v_{max} = u_x = 15 \, \text{m/s}
\]
### Final Answer
The velocity of the stone at the highest point is \(15 \, \text{m/s}\).
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