a stone is from the top of a tower 300m high and at the same time another stone is projected vertically upward with initial velocity 75m/s. calculate when and where the two stone meet

Solve for the t when

first stone height = second stone height
300 - 4.9 t^2 = 75 t- 4.9 t^2

The "4.9" is g/2, in m/s^2

75 t = 300
t = 4 seconds
height = 300 - 4.9*16 = 226 m

To find out when and where the two stones meet, we can start by analyzing the motion of each stone separately.

Let's consider the stones as Stone A (dropped from the tower) and Stone B (projected upward).

For Stone A:
- Initial velocity (u) = 0 m/s (as it is dropped)
- Acceleration (a) = 9.8 m/s² (acceleration due to gravity, acting downward)

For Stone B:
- Initial velocity (u) = 75 m/s (projected upward)
- Final velocity (v) = 0 m/s (at the highest point, where it changes direction)
- Acceleration (a) = -9.8 m/s² (acceleration due to gravity, acting downward)

We will use the kinematic equations to solve for the time and height at which the two stones meet.

For Stone A:
Since the initial velocity is zero, we can use the equation:

h = ut + (1/2)at²

Where:
h = height traveled (300m in this case)
u = initial velocity (0 m/s)
a = acceleration (-9.8 m/s²)
t = time

Using the given values, we get:

300m = 0 * t + (1/2) * (-9.8 m/s²) * t²

Simplifying:

300m = -4.9t²

Dividing both sides by -4.9:

t² = -300 / -4.9

t² ≈ 61.22

Finding the square root of both sides:

t ≈ √61.22

t ≈ 7.82 seconds (approx)

For Stone B:
Since the final velocity is zero, we can use the equation:

v = u + at

Where:
v = final velocity (0 m/s)
u = initial velocity (75 m/s)
a = acceleration (-9.8 m/s²)
t = time

Using the given values, we get:

0 = 75 + (-9.8) * t

Simplifying:

-75 = -9.8t

Dividing both sides by -9.8:

t = -75 / -9.8

t ≈ 7.65 seconds (approx)

Since Stone B reaches its highest point at approximately 7.65 seconds, and Stone A takes approximately 7.82 seconds to fall from the tower, we can conclude that the two stones meet around 7.65 seconds after Stone B is projected.

To find where they meet, we can calculate the height at that time by using the equation:

h = ut + (1/2)at²

Substituting the values for Stone B at t = 7.65s:

h = (75 m/s) * (7.65s) + (1/2) * (-9.8 m/s²) * (7.65s)²

Simplifying:

h ≈ 288.62 meters (approx)

Therefore, the two stones meet approximately 7.65 seconds after Stone B is projected, at a height of approximately 288.62 meters above the ground.