Question

Figure \(Q 1(b)\) shows a \(2 \mathrm{~kg}\) ball rolling down a glass ramp from \(A\) to \(B\). It is observed that the ball leaves the ramp at point \(B\) at an angle \(\theta=25^{\circ}\) with the horizontal. The height of the ramp is \(1.3 \mathrm{~m}\) from the floor. The ball has an initial velocity of \(V_{0}=2.5 \mathrm{~m} / \mathrm{s}\) and its speed is increasing at a rate of \(4.9 \mathrm{~m} / \mathrm{s}^{2}\). If the ball strikes the ground at D, determine how long takes to travel from \(\mathrm{A}\) to \(\mathrm{C}\). (CO1-PO1)(C4)

Answer 1

Let's calculate the time in the ramp: d = Votat? Using the giving data: 1.5 = 2.5 *t + 2.45t2 Solving we get: t = 0.426 s 1.7 The speed at the end of the ramp is: v = vo + at = 2.5 + 4.9 * 0.426 = 4.5874 The time in air (from B to C) Δx Δx 'Vxvcos(25) 4.5874 * cos(25) t = 0.409 s The total time is: T = ti + t2 = 0.835 s t = - =