Question

A stone is thrown from an elevation of 2 m above ground level, at an angle of \(30^\circ\) to the horizontal axis. If the stone hits the ground at a horizontal distance of 6 m from the point of release, at what speed (in m/s) was the stone thrown? Use \(g = 10\ \text{m/s}^2\) and assume that there is no air resistance. (Round off your answer to one decimal place.)

Hint:

- Break projectile motion into horizontal and vertical components. - Always use elimination of \(t\) to combine horizontal and vertical equations. - For problems with initial elevation, include initial height in vertical equation.

Answer 1

Correct Answer: 6.3

Step 1: Define initial velocity components. Let the initial speed be \(u\). \[ u_x = u\cos 30^\circ = \frac{\sqrt{3}}{2}u, \qquad u_y = u\sin 30^\circ = \frac{1}{2}u. \] Step 2: Equation of vertical motion. Vertical displacement after time \(t\): \[ y(t) = 2 + u_y t - \tfrac{1}{2}gt^2. \] At ground hit, \(y(t)=0\): \[ 0 = 2 + \tfrac{1}{2}u t - 5t^2. \tag{1} \] Step 3: Horizontal motion condition. Horizontal displacement after time \(t\): \[ x(t) = u_x t = \tfrac{\sqrt{3}}{2} u t = 6. \tag{2} \] Step 4: Express \(t\) in terms of \(u\). From (2): \[ t = \frac{12}{\sqrt{3}u} = \frac{4\sqrt{3}}{u}. \] Step 5: Substitute in vertical equation. From (1): \[ 0 = 2 + \tfrac{1}{2}u\left(\frac{4\sqrt{3}}{u}\right) - 5\left(\frac{4\sqrt{3}}{u}\right)^2. \] \[ 0 = 2 + 2\sqrt{3} - \frac{240}{u^2}. \] Step 6: Solve for \(u^2\). \[ \frac{240}{u^2} = 2+2\sqrt{3}. \] \[ u^2 = \frac{240}{2(1+\sqrt{3})} = \frac{120}{1+\sqrt{3}}. \] Multiply numerator and denominator by \((1-\sqrt{3})\): \[ u^2 = \frac{120(1-\sqrt{3})}{1-3} = -60(1-\sqrt{3}) = 60(\sqrt{3}-1). \] Numerically: \(\sqrt{3}\approx1.732\). \[ u^2 \approx 60(0.732) = 43.92 \Rightarrow u \approx 6.63\ \text{m/s}. \] Step 7: Recheck horizontal range. With \(u=6.63\), \(t=\tfrac{4\sqrt{3}}{6.63}\approx1.04\) s. Then \(x=u_x t = (6.63\cos30^\circ)(1.04)\approx(5.74)(1.04)=5.97\approx6\ \text{m}. \checkmark\) Thus \(u \approx 6.6\ \text{m/s}\). Final Answer: \(6.6\ \text{m/s}\)