Question

A stone is dropped by a person from The top of a building .which is 200m tall. At the same time. Another stone is thrown upwards. With a velocity of 50m/s.why the person standing at the roof of the building. Find the time after which two stones meet?

Note:Use silmultaneous equation

Answer 1

Answer:

Explanation:

Given A stone is dropped by a person from The top of a building .which is  200 m tall. At the same time. Another stone is thrown upwards. With a velocity of 50 m/s.why the person standing at the roof of the building. Find the time after which two stones meet?

• Let distance travelled by stone 1 be x and stone 2 be y.
• We know the equation of motion  
• S = u t + 1/2 at^2
• Stone 1            
• x = 0 x t^2 + 1/2  gt^2
• x = 1/2 gt^2
• Now stone 2
• y = 50 t + 1/2  (-g)t^2
• y = 50 t – 1/2 gt^2
• x + y = 200
• 1/2 gt^2 + 50 t – 1/2 gt^2 = 200
• t = 200 / 50
• t = 4 sec
• x = 1/2 x 9.8 x 16
• x = 78.4 m
• y = 200 – x
• y = 200 – 78.4
• y = 121.6 m
• The stones meet after 4 secs at a height of 121.6 m from bottom of tower.