Question 145: Motion in 1D

Question

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 = 18 s. What is the value of v?  
(Take g = 10 m/s2)

Accepted Answer

From the question, we can say
distance moved by 1
st
ball in 18 s = distance moved by 2
nd
ball in 12 s.
So, distance moved by 1
st
ball in 18 s
= $${1 \over 2} 	imes 10 	imes {18^2}$$ = 1620 m
and distance moved by 2
nd
ball in 12 s
= $$vt + {1 \over 2}g{t^2}$$
$$ 	herefore $$ 1620 = $$v\left( {12} ight) + {1 \over 2} 	imes 10{\left( {12} ight)^2}$$
$$ \Rightarrow $$ $$v$$ = 75 m/s

Answer

75 m/s

Answer

55 m/s

Answer

40 m/s

Answer

60 m/s

Motion in 1D

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 = 18 s. What is the value of v? (Take g = 10 m/s2)

Choose the correct answer: