a ball is dropped from height of tower 300m .after 1s another ball is dropped with 20m/s from tower .when and where they meet?

first ball: h=300-1/2 g t^2

second ball: h= 300-20(t-1)-1/2 g (t-1)^2

set the two equations equal, and solve for t. THat is when

put t into either equation to find where.

To determine when and where the two balls will meet, we can use the equations of motion.

The first ball is dropped from a height of 300m, and we can use the equation of motion for free fall:

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

where:
h = height
u = initial velocity (which is 0 in this case as it is dropped)
g = acceleration due to gravity (approximated as 9.8 m/s^2)
t = time

For the first ball:
Given h = 300m and u = 0, we can rearrange the equation to determine the time it takes to hit the ground:

300 = (1/2)(9.8)t^2
600 = 9.8t^2
61.22 = t^2
t ≈ 7.82s (taking the positive root)

The first ball will hit the ground after approximately 7.82 seconds.

Now, let's find where the second ball meets the first ball. The second ball is dropped after 1 second with an initial velocity of 20 m/s.

Using the equation for position:

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

where:
h = height
u = initial velocity
g = acceleration due to gravity
t = time

For the second ball:
Given u = 20 m/s and g = 9.8 m/s^2, and we need to determine h and t. We know that t = 7.82s - 1s = 6.82s (since the second ball is dropped 1 second later).

h = u * t + (1/2) * g * t^2
h = 20 * 6.82 + (1/2) * 9.8 * (6.82^2)
h ≈ 136.4m + 230.14m
h ≈ 366.54m

The second ball will meet the first ball at a height of approximately 366.54m, and it will meet after approximately 6.82 seconds.

Therefore, the two balls will meet approximately 6.82 seconds after the second ball is dropped, at a height of approximately 366.54m.

To find when and where the two balls meet, we can start by determining the time it takes for each ball to reach the ground.

For the first ball dropped from a height of 300m, we can use the equation of motion:

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

Where:
s = distance traveled (300m)
u = initial velocity (0 m/s, since it was dropped)
a = acceleration due to gravity (-9.8 m/s^2)
t = time

Plugging in the values, we get:

300 = 0 * t + (1/2) * (-9.8) * t^2

Simplifying gives us the equation:

4.9t^2 - 300 = 0

Using the quadratic formula, we can find the value of t when the first ball reaches the ground:

t = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 4.9, b = 0, and c = -300. Plugging in these values, we get:

t = (√(0^2 - 4 * 4.9 * -300)) / 2 * 4.9

Simplifying further gives us two values for t: t ≈ 7.89s and t ≈ -7.89s.

Since time cannot be negative, we can discard the negative value. Therefore, it takes approximately 7.89 seconds for the first ball to reach the ground.

Now let's calculate the height of the second ball dropped with an initial velocity of 20m/s after 1 second.

Using the equation of motion:

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

Where:
s = distance traveled
u = initial velocity (20m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (1s)

Plugging in the values, we have:

s = 20 * 1 + (1/2) * (-9.8) * 1^2

Simplifying gives us:

s = 10.1m

The second ball covers a distance of 10.1m in 1 second.

To determine where and when the two balls meet, we can subtract the height covered by the second ball from the height of the tower:

300m - 10.1m = 289.9m

So, the balls meet at a height of approximately 289.9m from the ground.

Since it took the first ball 7.89 seconds to reach the ground, we can calculate when they meet by subtracting 7.89 seconds from the time the second ball was dropped (1 second):

1s - 7.89s ≈ -6.89s

Therefore, they meet approximately 6.89 seconds before the second ball was dropped, or when the time was at -6.89 seconds.