Class 11th

Initial velocity, u₁ = 15m/s, acceleration, a = -g = -10 m/s

From the relation  s₁=s₀+u₁t+ (1/2)at² where

s₀ = cliff height, s₁ = total height of the fall of the first stone, we get

s₁ = 200 + 15t – 5t²  ………. (1)

When the stone hit the floor, s₁ = 0, so the equation (1) becomes

0 = 200 +15t - 5t² = t² -3t – 40 = (t-8) (t+5) = 0

So t = 8s or -5s

Since the stone was thrown at t=0, so t cannot be –ve. Hence t = 8s

For the second stone,

Initial velocity, u₁ = 30 m/s, acceleration, a = -g = -10 m/s

From the relation  s₂=s₀+u₁t+ (1/2)at² where

s₀ = cliff height, s₂= total height of the fall of the second stone, we get

s₂ = 200 + 30t – 5t²  ………. (2)

When the stone hit the floor, s₂ = 0, so the equation (2) becomes

0 = 200 +30t - 5t² = t² -6t – 40 = (t-10) (t+4) = 0

So t = 10s or -4s

Since the stone was thrown at t=0, so t cannot be –ve. Hence t = 10s

Subtracting equation (1) from equation (2), we get

s_{2 –} s₁ = (200 + 30t – 5t²  ) – (200 + 15t – 5t² )  = 15t ……. (3)

Equation (3) represents the linear trajectory of the two stones because to this linear relation between (s_{2 –} s₁₎ and t, the projection is straight line till 8s (from the graph)

The maximum distance between the two stones is at t = 8s. So

(s_{2 –} s₁) _max = 15 x 8 = 120 m. This value has been depicted correctly in the graph.

After 8s, only the second stone is in motion (t for first stone has been found out to be 8s), the trajectory of second stone is obtained in equation (2), = 200 + 30t – 5t²  , which is a curved path.

3.23 The distance covered by the 3 wheeler on a straight line in the nth second can be expressed as:

Sn = u + a (2n-1)/2 …… (1),

Where

a = acceleration

u = initial velocity

n = time = 1,2,3, ……., n

Given, u = 0, a = 1m/s², from equation (1) we get Sn = (2n-1)/2 ……. (2)

With various values of n, we get Sn

            n                      Sn

            1                      0.5

            2                      1.5

            3                      2.5

            4                      3.5

            5                      4.5

            6                      5.5

            7                      6.5

            8                      7.5

            9                      8.5

            10                    9.5

(a) This graph is a Displacement-Time Graph (x- t) that shows the displacement increasing From A to B, and decreasing From B to C. The graph may represent a carom board where the striker hits the edge, rebounds with reduced speed, then moves in the opposite direction, hits the opposite wall and stops.

(b) This graph is a Velocity-Time Graph (v- t) that shows Sudden spikes and drops in velocity. There is a rapid change in velocity. This graph may represent a ball that falls on the ground from a certain height and rebounds with a reduced speed

(c) This graph is an Acceleration -Time Graph (a- t) that shows an instantaneous spike in acceleration. The graph looks like of a baseball, thrown on to a batter, moves with a uniform speed and then hit by the bat for a very short time interval.