Question

Stone dropped from top of a tower 100m high. At same instant another stone is thrown vertically from base of the tower with a velocity of 25m/s. When and where will the two stones meet? Given g =10 m/s^2

Answer 1

see other post.

Answer 2

let the height of the dropped stone be s1

s1 = ut + 0.5gt^2
u=0
s1 = 0.5*10*t^2 = 5t^2

let the height of the thrown stone be s2
s2 = ut - 0.5gt^2
s2 = 25t - 5t^2

s1+s2 = 100
5t^2 + 25t - 5t^2 = 100
25t = 100
t = 4s

the stones meet 4seconds later

s2 = 25t - 5t^2 = 100 - 5(16)
s2 = 100 - 80 = 20m

the stones meet 20m from the base of the tower.

Answer 3

Thanks helped me a lot

Answer 4

very helpful. thanx a lot

Answer 5

[AI-generated answer — Explain Bot]

To find when and where the two stones meet, we need to determine the time it takes for each stone to reach the same height.

First, let's consider the stone dropped from the top of the tower. We can use the equation for free fall motion:

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

where:
h = height (100m in this case)
u = initial velocity (0m/s because the stone is dropped)
g = acceleration due to gravity (-10m/s^2 because it is directed downward)
t = time taken

Plugging in the values, we get:

100 = 0*t + (1/2)*(-10)*t^2
100 = -5t^2

Rearranging the equation, we have:

5t^2 = -100

Dividing both sides by 5, we get:

t^2 = -20

This equation has no real solutions, so it means that the stone dropped from the top of the tower will not meet the stone thrown from the base of the tower.

Now let's consider the stone thrown from the base of the tower:
We can use the same equation for this stone but with a positive initial velocity of 25m/s:

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

Plugging in the values, we get:

h = 25*t + (1/2)*(-10)*t^2
h = 25t - 5t^2

Since we want to know where the two stones meet, we set the heights equal to each other:

25t - 5t^2 = 100

Rearranging the equation, we have:

5t^2 - 25t + 100 = 0

We can solve this quadratic equation for t using factoring, completing the square, or using the quadratic formula. In this case, let's use factoring:

(5t - 20)(t - 5) = 0

This gives us two solutions:

5t - 20 = 0 --> t = 4s
t - 5 = 0 --> t = 5s

Therefore, the two stones will meet at t = 4s and t = 5s.

To find where they meet, we can substitute the values of t back into the equation for the height:

When t = 4s:
h = 25t - 5t^2
h = 25*4 - 5*4^2
h = 100 - 80
h = 20m

Therefore, the two stones will meet at a height of 20m above the base of the tower, 4 seconds after the stone is thrown from the base.