A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18s. What is the value of v?
Nipukalita:
135m/s
Answers
Answered by
2
the distance covered by the first ball in 18s=18²g/2
=162×10
the distance covered by the second ball=vt+1/2gt²
=12v+1/2×10×12²
=12v+5×144
=12v+720
12v+720=1620
v=900/2=450
=162×10
the distance covered by the second ball=vt+1/2gt²
=12v+1/2×10×12²
=12v+5×144
=12v+720
12v+720=1620
v=900/2=450
Distance covered by the first ball in 18 s,
Answered by
5
Distance covered by the first ball in 18 s,
S=1/2×10×18×18
=1620m
Now for the 2nd ball to cover 1620 m in 12 s, its velocity will be
=1620/12
=135m/s
S=1/2×10×18×18
=1620m
Now for the 2nd ball to cover 1620 m in 12 s, its velocity will be
=1620/12
=135m/s
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