Problem Statement: A ball is dropped from the top of a tower and 1 second later another ball is thrown vertically downward with an initial velocity of 15m/s. Find the height of the tower if the two balls strike the ground simultaneously.

Solution:

Step 1: Calculate the time taken by the first ball to reach the ground.

The time taken by the first ball to reach the ground can be calculated using the equation:

h = 1/2 * g * t^2

where h is the height of the tower, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken by the first ball to reach the ground.

Rearranging the above equation, we get:

t = sqrt(2h/g)

Substituting the values, we get:

t = sqrt(2h/9.8)

t = sqrt(h/4.9)

Since the second ball is thrown 1 second after the first ball is dropped, the time taken by the second ball to reach the ground is t - 1.

Therefore, the time taken by the second ball to reach the ground is:

t - 1 = sqrt(h/4.9) - 1

Step 2: Calculate the height of the tower.

Since both balls strike the ground simultaneously, the time taken by the first ball to reach the ground is equal to the time taken by the second ball to reach the ground.

Therefore, we can equate the two expressions for time and solve for h:

sqrt(h/4.9) = sqrt(h/9.8) - 1

Squaring both sides, we get:

h/4.9 = h/9.8 - 2*sqrt(h/9.8) + 1

Simplifying the above equation, we get:

2*sqrt(h/9.8) = h/4.9 - h/9.8 + 1

2*sqrt(h/9.8) = h/9.8 + 1

Squaring both sides, we get:

4*h/9.8 = h/9.8 + 1 + 2*sqrt(h/9.8)

3*h/9.8 - 1 = 2*sqrt(h/9.8)

Squaring both sides, we get:

9*h^2/96.04 - 3*h/9.8 + 1 = 4*h/9.8

9*h^2/96.04 - 7*h/9.8 + 1 = 0

Solving the above quadratic equation, we get:

h = 49.05 m

Therefore, the height of the tower is 49.05 m.

Conclusion: The height of the tower is 49.05 m.