Answer:

Explanation:

Q.1:

Let's consider ball X first. The time taken by ball X to reach the ground is 6 seconds. We know that the time of flight for an object thrown vertically upwards and then coming back to the ground is twice the time taken to reach the highest point. So, the time taken by ball X to reach the highest point is 6/2 = 3 seconds.

Using the equation of motion:

v = u - gt,

where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (10 m/s²), and t is the time taken.

At the highest point, the final velocity v becomes 0, so we can write:

0 = u - gt.

Substituting the values, we have:

0 = u - 10 * 3.

Simplifying the equation, we get:

u = 30 m/s.

Therefore, the initial speed of ball X is 30 m/s.

Now let's move on to ball Y. The time taken by ball Y to reach the ground is given as 2 seconds. Since ball Y is thrown vertically downward, the initial velocity will be negative. Using the equation of motion:

h = ut + (1/2)gt²,

where h is the height of the tower, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.

At the ground level, the height h becomes 0, so we can write:

0 = u * 2 + (1/2) * 10 * 2².

Simplifying the equation, we get:

0 = 2u + 20.

Solving for u, we have:

u = -10 m/s.

Therefore, the initial speed of ball Y is -10 m/s (negative indicating downward direction).

To find the height of the tower, we can use the equation:

h = ut + (1/2)gt².

Substituting the values for ball X, we have:

0 = 30 * 3 + (1/2) * 10 * 3².

Simplifying the equation, we get:

0 = 90 + 45.

Since the equation is not satisfied, there might be an error or missing information in the question.

Q.2:

To answer the questions regarding the motion of ball A and ball B, more information is needed. Specifically, the initial speed u and the height h need to be provided. With these additional details, I can help you calculate the separation between the bodies at any time instant, the time instant at which they collide, and the position of ball A at the moment of collision.

Verified answer

Q.1 Two balls X and Y are thrown from top of tower one vertically upward and other vertically downward with same speed. If times taken by them to reach the ground are 6 s and 2 s respectively, then the height of the tower and initial speed of each ball are (g = 10 m/s²)

Paragraph for Q. Nos. 6 to 11

Q.2 A ball A is thrown up vertically with speed u. At the same instant another ball B is released from rest at a height h vertically above the point of projection of A. Taking the moment of projection as t= 0 and acceleration due to gravity is 'g, answer the following questions.

1. The separation between the bodies at any time instant t is

2.The time instant at which they collide is

3. At the moment of collision, the position of A is at a distance

Q.1: To find the height of the tower and the initial speed of the balls, we can use the equations of motion.

Let's consider the ball thrown upwards as ball X and the ball thrown downwards as ball Y.

For ball X:

Using the equation s = ut + (1/2)gt^2, where s is the distance covered, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.

Here, s = height of the tower, u = initial speed of ball X, g = 10 m/s², and t = 6 s.

Plugging in the values, we get:

s = ut + (1/2)gt^2

s = 6u - 180 ...equation (1)

For ball Y:

Using the same equation, s = ut + (1/2)gt^2, where s is the distance covered, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.

Here, s = height of the tower, u = initial speed of ball Y, g = 10 m/s², and t = 2 s.

Plugging in the values, we get:

s = ut + (1/2)gt^2

s = 2u - 20 ...equation (2)

To find the height of the tower, we can equate equations (1) and (2):

6u - 180 = 2u - 20

Simplifying the equation:

4u = 160

u = 40 m/s

Substituting the value of u in equation (1):

s = 6u - 180

s = 6 * 40 - 180

s = 240 - 180

s = 60 meters

Therefore, the height of the tower is 60 meters and the initial speed of each ball is 40 m/s.

Q2/ Let's solve each question step by step:

The separation between the bodies at any time instant t is:

To determine the separation between the two balls at any time instant, we need to consider their respective positions at that time.

Ball A's position at any time t can be calculated using the kinematic equation for vertical motion:

s_A = u*t - (1/2)gt^2

Ball B's position at any time t can also be calculated using the same equation:

s_B = h - (1/2)gt^2

The separation between the two balls at any time t is the difference between their positions:

separation = s_B - s_A

separation = h - (1/2)gt^2 - (ut - (1/2)gt^2)

separation = h - ut

Therefore, the separation between the balls at any time t is h - u*t.

The time instant at which they collide is:

For the two balls to collide, their positions must be equal at the same time. We can set up an equation to find the time at which this occurs.

Setting s_A = s_B, we have:

u*t - (1/2)gt^2 = h - (1/2)gt^2

Simplifying and rearranging the equation:

u*t = h

t = h/u

Therefore, the time instant at which the balls collide is t = h/u.

At the moment of collision, the position of A is at a distance:

To find the position of ball A at the moment of collision, we substitute the collision time (t = h/u) into the equation for ball A's position:

s_A = ut - (1/2)gt^2

s_A = u(h/u) - (1/2)g(h/u)^2

s_A = h - (1/2)g(h^2/u)

Therefore, at the moment of collision, the position of ball A is at a distance of h - (1/2)g(h^2/u).