From the foot of a tower 90m high a stone is thrown up so as to reach the top of the tower .two second later another stone is dropped from the top of the tower when and where the stone will maeet
Answers
Height of the building (d) = 90m
Let’s find the velocity of stone thrown up:
We know, v^2 – u^2= 2ad
Velocity of stone after reaching top of the building = 0
Acceleration is opposite to acceleration due to gravity = -9.8m/s^2
0- u^2= 2*(-9.8)*90
- u^2=-1764
Therefore, velocity of stone moving up= 42m/s
Distance at time t = h(t) =ut-1/2at^2
h(t) = 42t-4.9t^2 -----(1)
h1(T) = height of building – 1/2aT^2
T=t-2 (since second stone is thrown after two seconds)
H1= 90 – 4.9(t-2)^2
When the two stones meet, distance are same. So, Set h= h1
42t - 4.9t^2 = 90 -4.9t^2 +19.6t -19.6
42t-19.6t=90-19.6
22.4t = 70.4 => t = 3.14s
Substitute the value of “t” in equation (1)
h =83.6m
So, the two stones meet at 83.6m
Answer:
Height of the building (d) = 90m
Let’s find the velocity of stone thrown up:
We know, v^2 – u^2= 2ad
Velocity of stone after reaching top of the building = 0
Acceleration is opposite to acceleration due to gravity = -9.8m/s^2
0- u^2= 2*(-9.8)*90
- u^2=-1764
Therefore, velocity of stone moving up= 42m/s
Distance at time t = h(t) =ut-1/2at^2
h(t) = 42t-4.9t^2 -----(1)
h1(T) = height of building – 1/2aT^2
T=t-2 (since second stone is thrown after two seconds)
H1= 90 – 4.9(t-2)^2
When the two stones meet, distance are same. So, Set h= h1
42t - 4.9t^2 = 90 -4.9t^2 +19.6t -19.6
42t-19.6t=90-19.6
22.4t = 70.4 => t = 3.14s
Substitute the value of “t” in equation (1)
h =83.6m
So, the two stones meet at 83.6m
Step-by-step explanation: