Question

A stone is dropped from a top of tower of height 490m simultaneously another stone is projected up vertically with velocity 100m/a from the ground the two stones meet after a time?

Answer 1

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The height from the ground of stone 1 dropped from the tower after t seconds:

h1=490-gt^2/2 =490–4.9t^2

The height of stone 2 thrown upwards at 100 m/s after t seconds:

h2 = v0t-gt^2/2= 100t-4.9t^2

The two stones will meet when:

h1 = h2

490–4.9t^2 = 100t -4.9t^2

100t = 490

t =490/100 = 4.9 s

The two stones will meet after 4.9 seconds

Answer 2

Answer

h=490 m U=100

we have a formula that t=h/

Step-by-step explanation:

so answer is 4.9