A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from same platform with speed v. The two balls meet at t= 18 second. What is the value of v?
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Answer:
75 m/s
Step-by-step explanation:
Distance of 1st ball= S= 1/2 x 10 x 18
= 1620
Distance of 2nd ball = S' = ut + 1/2gt²
1620 = 12u + 5 x 144
u = 135 - 60
= 75 m/s
bishakha84:
I didn't understand
Okay now I got it
mark me thanks atleast
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