A stone piece is dropped from the top of a tower of height 90 m and another stone piece is thrown vertically up words with a speed 30 m/s at the same time. When and where the stone pieces will meet?
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Answer:
Let the two stones meet at a distance of x mfrom the top of the tower, and 't' be the time taken. Let us assume the downward direction as positive.
For the stone that is dropped its initial velocity u=0ms
−1
; displacement s = x and acceleration = acceleration due to gravity (g).
Using s=ut+
2
1
at
2
,
wegetx=(0)t+
2
1
gt
2
→(1).
For the stone that is projected vertically upwards, its initial velocity, u=−50ms
−1
; displacement s =-(200-x) and acceleration a =g.
using s=ut+
2
1
at
2
,
weget−(200−x)=−50×t+1/2gt
2
200=50t−1/2gt
2
+x→(1)(2)
Fromtheequations(1)and(2)
wehave200=50t−1/2gt
2
+1/2gt
2
⇒200=50∴t=4s
Explanation:
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