Two stones are projected vertically up wards one after the other with same velocities 20 m s1 with a time interval of 2 sec. After what time and at what height from point of projection do they meet?

To solve the problem of when and where the two stones meet after being projected vertically upwards, we can break it down into a few logical steps. Both stones are projected with the same initial velocity of 20 m/s, but they are launched 2 seconds apart. Let's analyze their motions using the equations of motion under uniform acceleration due to gravity.

Understanding the Motion of the Stones

When an object is projected upwards, it experiences a downward acceleration due to gravity, which we can approximate as 9.81 m/s² (we'll use 10 m/s² for simplicity in calculations). The height \( h \) of an object at any time \( t \) can be calculated using the formula:

h = ut - (1/2)gt²

Where:

• u = initial velocity (20 m/s)
• g = acceleration due to gravity (10 m/s²)
• t = time in seconds

Calculating the Heights of Each Stone

Let’s denote the first stone as Stone A and the second stone as Stone B. Stone A is projected at \( t = 0 \) seconds, and Stone B is projected at \( t = 2 \) seconds.

Height of Stone A

For Stone A, the height at any time \( t \) is given by:

h_A = 20t - 5t²

Height of Stone B

Stone B is launched 2 seconds later, so when it is launched, Stone A has already been in motion for 2 seconds. Therefore, the time for Stone B is \( t - 2 \) seconds. The height of Stone B can be expressed as:

h_B = 20(t - 2) - 5(t - 2)²

Setting the Heights Equal

To find the time when both stones meet, we set their heights equal to each other:

20t - 5t² = 20(t - 2) - 5(t - 2)²

Expanding the right side:

20t - 5t² = 20t - 40 - 5(t² - 4t + 4)

20t - 5t² = 20t - 40 - 5t² + 20t - 20

Now simplifying:

20t - 5t² = 40t - 60 - 5t²

Canceling \( -5t² \) from both sides gives:

0 = 20t - 60

Solving for \( t \):

20t = 60

t = 3 seconds

Finding the Height at Which They Meet

Now that we have the time at which they meet, we can find the height by substituting \( t = 3 \) seconds back into the height equation for either stone. Using Stone A's equation:

h_A = 20(3) - 5(3)²

h_A = 60 - 45 = 15 meters

Final Results

In summary, the two stones meet after 3 seconds at a height of 15 meters from the point of projection. This problem illustrates the principles of kinematics and how to analyze the motion of multiple objects under the influence of gravity.