(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 vertical...

(a) Uniform acceleration is motion in which the velocity changes by equal amounts in equal intervals of time. Thus, the rate of change of velocity is constant.

(b)(i) Taking upward velocity as positive, the velocity-time graph is as shown below. It is a straight line of constant negative gradient. The ball starts with velocity u, slows uniformly to zero velocity at the highest point, and then gains downward velocity uniformly until it returns to the point of projection with velocity −u.

(b)(ii) The gradient of a velocity-time graph gives acceleration:

\[\text{gradient}=\frac{\Delta v}{\Delta t}=-g.\]

Hence the acceleration due to gravity is the magnitude of the gradient:

\[g=\left|\frac{\Delta v}{\Delta t}\right|.\]

The maximum height attained is the area under the graph above the time axis, from projection to the instant the velocity becomes zero. Since this area is a triangle,

\[H=\frac{1}{2}\times u\times t_{\rm up}.\]

Since \(t_{\rm up}=u/g\),

\[H=\frac{1}{2}u\left(\frac{u}{g}\right)=\frac{u^2}{2g}.\]

The triangular area below the time axis during the downward journey has the same magnitude, showing that the ball falls through the same vertical distance before returning to the point of projection.

(c) Take upward as positive and \(g=10\,\text{m s}^{-2}\).

(i) Let \(t\) be the time for which the second stone is in motion before the stones meet. For the second stone, \(u=20\,\text{m s}^{-1}\) and \(v=10\,\text{m s}^{-1}\):

\[v=u-gt\]

\[10=20-10t\]

\[10t=10\]

\[t=1\,\text{s}.\]

Therefore, the second stone has been in motion for \(1\,\text{s}\).

(ii) The first stone has been in motion for \(2+1=3\,\text{s}\). Therefore,

\[v=u-gt=20-(10\times3)=-10\,\text{m s}^{-1}.\]

The velocity of the first stone is \(-10\,\text{m s}^{-1}\), that is, \(10\,\text{m s}^{-1}\) vertically downward.