(a) Explain the term uniform acceleration
(b)(i) Sketch and describe the velocity-time graph for the motion of a ball from the time it is projected vertically upwards until it returns to the point of projection.
(ii) Neglecting air resistance and using ycur sketch, explain how the acceleration of free fall due to gravity g, and the maximum height attained when the ball is projected vertically upwards can be determined.
(c) A stone is projected vertically upwards with a velocity of 20ms\(^{-1}\). Two seconds later, a second stone is similarly projected with the same velocity. When the two stones meet, the second one is rising at a velocity of 10ms\(^{-1}\). Neglecting air resistance, calculate the:
(i) length of time the second stone is in motion before they meet,
(ii) velocity of the first stone when they meet (Take g as 10ms\(^{-2}\))
(a) Uniform acceleration is motion in which the velocity of a body changes by equal amounts in equal intervals of time; that is, the rate of change of velocity is constant.
(b)(i) The velocity-time graph is a straight line sloping downward from the initial upward velocity u (positive) to zero at the highest point, then continuing at the same slope into negative values until the ball returns to the point of projection with velocity -u (equal magnitude, opposite direction). The graph is a single straight line of constant negative slope crossing the time axis at the top of the flight.
(b)(ii) The slope (gradient) of the line is constant and equal in magnitude to g; therefore \(g\) is obtained as the magnitude of the gradient \(\left|\dfrac{\Delta v}{\Delta t}\right|\). The maximum height is the area between the graph and the time axis during the upward journey (a triangle), \(H=\tfrac{1}{2}\times t_{up}\times u\), where \(t_{up}\) is the time to reach the highest point.
(c) Let the first stone start at t = 0 with u = 20 m/s. The second starts 2 s later. Taking upward as positive with g = 10 m/s\(^2\).
(i) For the second stone at the meeting instant it is rising at 10 m/s: \(v=u-gt_2\Rightarrow10=20-10t_2\Rightarrow t_2=1\,\text{s}\). The second stone has been in motion 1 s before they meet.
(ii) The first stone has then been moving for \(t_1=t_2+2=3\,\text{s}\). \(v_1=u-gt_1=20-10(3)=-10\,\text{m/s}\). So the first stone is moving downward at 10 m/s. (Check: heights are \(20(3)-5(3)^2=15\,\text{m}\) and \(20(1)-5(1)^2=15\,\text{m}\); they indeed meet.)
(a) Uniform acceleration is motion in which the velocity of a body changes by equal amounts in equal intervals of time; that is, the rate of change of velocity is constant.
(b)(i) The velocity-time graph is a straight line sloping downward from the initial upward velocity u (positive) to zero at the highest point, then continuing at the same slope into negative values until the ball returns to the point of projection with velocity -u (equal magnitude, opposite direction). The graph is a single straight line of constant negative slope crossing the time axis at the top of the flight.
(b)(ii) The slope (gradient) of the line is constant and equal in magnitude to g; therefore \(g\) is obtained as the magnitude of the gradient \(\left|\dfrac{\Delta v}{\Delta t}\right|\). The maximum height is the area between the graph and the time axis during the upward journey (a triangle), \(H=\tfrac{1}{2}\times t_{up}\times u\), where \(t_{up}\) is the time to reach the highest point.
(c) Let the first stone start at t = 0 with u = 20 m/s. The second starts 2 s later. Taking upward as positive with g = 10 m/s\(^2\).
(i) For the second stone at the meeting instant it is rising at 10 m/s: \(v=u-gt_2\Rightarrow10=20-10t_2\Rightarrow t_2=1\,\text{s}\). The second stone has been in motion 1 s before they meet.
(ii) The first stone has then been moving for \(t_1=t_2+2=3\,\text{s}\). \(v_1=u-gt_1=20-10(3)=-10\,\text{m/s}\). So the first stone is moving downward at 10 m/s. (Check: heights are \(20(3)-5(3)^2=15\,\text{m}\) and \(20(1)-5(1)^2=15\,\text{m}\); they indeed meet.)
