A stone is dropped from top of the tower 400 m, and at the same time another stone is projected upward vertically from the ground with velocity of 100 m/s. find where and when the two stones will meet.
Answers
Answer:
320 m from the ground or 80 m from the top. Time when they meet = 4s
Explanation:
Consider the first stone. For the first stone:-
u₁ = 0 m/s
a₁ = g = 10 m/s²
For the second stone,
u₂ = 100 m/s
a₂ = - g = - 10 m/s² (negative because acceleration due to gravity will be in opposite direction of the motion)
Now, let the distance at which the two stones meet be xm from the ground. So, the second stone would travel x m and the first stone would travel (400 - x)m as for the second stone, the distance would be considered from the top of the tower (refer the attachment).
So, for the first stone,
s₁ = u₁t + 1/2 a₁t²
⇒ (400 - x) = 1/2 × 10t² (since, u = 0 m/s)
⇒ (400 - x) = 5t² ...................(1st equation)
Now, for the second stone,
s₂ = u₂t + 1/2a₂t²
⇒ x = 100t - 1/2 × 10t² (since, a₂ = - 10)
⇒ x = 100t - 5t² ....................(2nd equation)
Now, we know that 5t² = (400 - x) [from the first equation]. So put 5t² as 400 - x in 2nd equation.
⇒ x = 100t - (400 - x)
⇒ x = 100t - 400 + x
⇒ x - x = 100t - 400
⇒ 0 = 100t - 400
⇒ 400 = 100t
⇒ t = 400/100
⇒ t = 4 seconds
Now x = 100t - 5t²
⇒ x = 100(4) - 5(4)²
⇒ x = 400 - 5(16)
⇒ x = 400 - 80
⇒ x = 320 metres
Now, x is the distance measured from the ground.
Hence, the two stones would meet at a distance of 320 m from the ground at time = 4 seconds.
Motion :
Answer :
The stones will meet at 321.6 m above the ground and at 4 second.
Explanation :
Refer the attached picture.
