Question

A rubber ball is released from a height about 1.5 m. It is caught after three bounces. Sketch graphs of its position, velocity, and acceleration as functions of time. Take positive-direction as the upward direction.

Answer 1

Check the following points:
a.Position and time graphs will have the same shape as the given graph.

b.The slope of the position-time graph gives the velocity. Hence, changes in velocity can be observed as per the slope of the position-time graph.

c.The slope of the velocity-time graph gives the acceleration.

d.Acceleration in free fall remains constant with time.

second stone is released.

Sol. Let at any time t = 0, the balloon be at position A, where its velocity is u (Fig. 4.13 8). At /=2 s, it reaches B, where its velocity becomes v, then solve to get x= 55 m.

Alternatively: (This method can be understood in the proper way after studying relative velocity.)
If we work from the frame of a balloon, then the acceleration of each stone w.r.t. a balloon will be g + a after releasing from it. The initial velocity of each stone will be zero w.r.t. balloon.
S,= 1 ( + fl)(3.5)2,S2 = i(g + a)(1.5)2;* = 5, -S2 = 55mExample 4.3 A rubber ball is released from a height about 1.5 m. It is caught after three bounces. Sketch graphs of its position, velocity, and acceleration as functions of time. Take positive-direction as the upward direction. Sol. Check the following points:
a. Position and time graphs will have the same shape as the given graph.
b.The slope of the position-time graph gives the velocity. Hence, changes in velocity can be observed as per the slope of the position-time graph.
c.The slope of the velocity-time graph gives the acceleration.
d.Acceleration in free fall remains constant with time. Note: Velocity is changing at points B, C, D, E, F. At B, D, and F, velocity changes suddenly from negative to positive and at C and E, velocity changes smoothly from positive to negative.
At B, D and F, velocity changes very quickly so the acceleration must be very large.
Position-time graph: The diagram is given itself conveys the position-time graph.
velocity-time graph: As the slope of the v-t graph changes thrice negative to positive during bounces. So v-t graph must observe sharp changes at these points.
Acceleration-time graph: Slope of v-t graph remains the same (-10 $m{s}^{-2}$), i.e., it has single value during free fall. At the time of contact with floor changes substantially during a very short time interval. So the acceleration will be large and positive, which straight upward lines.