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Step-by-Step Detailed Solution
Step 1: Identify the Known Parameters
• Initial height from the ground, $h = 10\text{ m}$.
• Initial speed of the ball, $v_0 = 10\text{ m/s}$.
• Angle of projection with the horizontal, $\theta = 30^\circ$.
• Acceleration due to gravity, $g = 10\text{ m/s}^2$.
• $\sin 30^\circ = \frac{1}{2},\;\cos 30^\circ = \frac{\sqrt{3}}{2}$.
Step 2: Resolve the Velocity into Components
The initial velocity has two components:
Vertical component of initial velocity:
$v_{0y} = v_0 \sin \theta = 10 \times \frac{1}{2} = 5\text{ m/s}$.
Horizontal component of initial velocity:
$v_{0x} = v_0 \cos \theta = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\text{ m/s}$.
Step 3: Write the Equation for Vertical Motion
Let the ground be the reference (i.e., $y=0$). The ball is initially at $y_0=10\text{ m}$. The vertical position at any time $t$ is:
$y(t) = y_0 + v_{0y}t - \frac{1}{2}\,g\,t^2.$
Substituting the known values:
$y(t) = 10 + (5)t - \frac{1}{2} (10) t^2 = 10 + 5t - 5t^2.$
Step 4: Solve for the Time when the Ball is at 10 m Again
We want $y(t) = 10\text{ m}$ (the same height as the throwing point). So,
$10 + 5t - 5t^2 = 10.$
Simplifying,
$5t - 5t^2 = 0,$
$5t(1 - t) = 0 \implies t = 0 \text{ or } t = 1\text{ s}.$
$t=0$ corresponds to the initial moment. Hence, the later time when the ball returns to the height of 10 m from the ground is $t=1\text{ s}$.
Step 5: Calculate the Horizontal Distance Covered in this Time
The horizontal velocity remains constant (neglecting air resistance). Thus,
$x(t) = v_{0x}\, t = 5\sqrt{3} \times 1 = 5\sqrt{3}\text{ m}.$
Numerically, $\sqrt{3} \approx 1.732,$ so
$5 \times 1.732 = 8.66\text{ m}.$
Step 6: State the Final Answer
Hence, the ball will be at a horizontal distance of 8.66 m from the point of projection when it is again at the height of 10 m from the ground.
