a stone A is dropped from the top of tower and B is thrown downward from same point 2 seconds later with initial velocity of 30 ms , if the same stone A and B reach the ground at same time, find height of the tower

da = 0.5g*t^2 = 4.9*2^2=19.6 m. in 2 s.

Va = Vo + g*t = 0 + 9.8*2 = 19.6 m/s in
2 s.

Da = Db - 19.6
19.6t + 4.9t^2 = 30t + 4.9t^2 - 19.6
19.6t - 30t + 4.9t^2 - 4.9t^2 = -19.6
-10.4t = -19.6
t = 1.88 s. to reach gnd.

h = 30t + 4.9t^2 = 30*1.88 + 4.9*1.88^2
= 73.9 m.

To find the height of the tower, we can use the equations of motion. Let's break down the problem into two phases: the motion of stone A and the motion of stone B.

For stone A:
1. We know that stone A is dropped from the top of the tower, so its initial velocity is 0 m/s.
2. We can use the equation of motion to find the height of the tower: h = (1/2) * g * t^2, where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time stone A takes to reach the ground.
3. We need to find the time it takes for stone A to reach the ground. Since it is dropped, we can use the equation t = sqrt(2h/g), where h is the height of the tower.
4. Substituting the known values, we get t = sqrt(2h/9.8).

For stone B:
1. Stone B is thrown downward from the same point, so its initial velocity is -30 m/s (downward).
2. Stone B starts 2 seconds after stone A, so its time of flight will be 2 seconds less than stone A.

Now, by the given condition of both stones hitting the ground at the same time, we can equate the times of flight for both stones: sqrt(2h/9.8) = sqrt((2h-30*2)/9.8).

Squaring both sides of the equation, we get:
2h/9.8 = (2h-60)/9.8.

Simplifying the equation:
2h = 2h - 60.

Hence, h = 60 meters.

Therefore, the height of the tower is 60 meters.