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Step-by-Step Solution
Step 1: Identify the given data
• Initial speed: $u = 10\,\text{m/s}$
• Angle of projection: $60^\circ$ with the horizontal
• Gravitational acceleration: $g = 10\,\text{m/s}^2$
• Time at which radius of curvature is calculated: $t = 1\,\text{s}$
Step 2: Resolve the initial velocity into components
The initial velocity can be broken into horizontal ($u_x$) and vertical ($u_y$) components:
\[
u_x = u\,\cos(60^\circ) = 10 \times \frac{1}{2} = 5\,\text{m/s}\\
u_y = u\,\sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\,\text{m/s}
\]
Step 3: Find the velocity components at $t = 1\,\text{s}$
Horizontal velocity remains constant (no horizontal acceleration):
\[
v_x = u_x = 5\,\text{m/s}
\]
Vertical velocity after time $t$ is given by
\[
v_y = u_y - g\,t = 5\sqrt{3} - 10 \,\text{m/s}
\]
Step 4: Calculate the magnitude of the velocity at $t = 1\,\text{s}$
\[
v = \sqrt{v_x^2 + v_y^2} \\
v = \sqrt{(5)^2 + (5\sqrt{3} - 10)^2}
\]
Numerically, $5\sqrt{3} \approx 8.66$, so
\[
v_y \approx 8.66 - 10 = -1.34\,\text{m/s}
\]
Therefore,
\[
v \approx \sqrt{25 + (-1.34)^2} \approx \sqrt{25 + 1.7956} \approx \sqrt{26.7956} \approx 5.18\,\text{m/s}
\]
Step 5: Determine the normal component of acceleration
The acceleration vector is $\mathbf{a}=(0,\,-10)$ and the velocity vector at $t=1\,\text{s}$ is $\mathbf{v}=(5,\,-1.34)$. The normal acceleration $a_n$ is the component of $\mathbf{a}$ perpendicular to $\mathbf{v}$, which can be found using the magnitude of the cross product over the velocity magnitude:
\[
a_n = \frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|}
\]
In 2D,
\[
\mathbf{v} \times \mathbf{a} = v_x \,a_y - v_y \,a_x = (5)(-10) - (-1.34)(0) = -50
\]
Hence the magnitude
\[
|\mathbf{v} \times \mathbf{a}| = 50
\]
So
\[
a_n = \frac{50}{|\mathbf{v}|} \approx \frac{50}{5.18} \approx 9.65\,\text{m/s}^2
\]
Step 6: Calculate the radius of curvature
The radius of curvature $R$ at any instant is given by
\[
R = \frac{v^2}{a_n}
\]
Substituting the values,
\[
R \approx \frac{(5.18)^2}{9.65} \approx \frac{26.8324}{9.65} \approx 2.78\,\text{m}
\]
Rounding suitably,
\[
R \approx 2.8\,\text{m}
\]
Final Answer
2.8 m
