Explanation
Let after a time t, the cyclist overtake the bus. Then
96+12×2×t2=20×t or t2−20t+96=0
∴t=20±√400−4×962×1
=20±42=8sec and 12sec
But since they won't meet again after they have crossed each other once, so only 8 seconds is the answer since they are headed in different directions.
Let a be the constant acceleration of the particle. Thens=ut+12at2 or s1=0+12×a×(10)2=50aand s2=[0+12a(20)2]−50a=150a
∴s2=3s1
Alternatively:Let a be constant acceleration and
s=ut+12at2, then s1=0+12×a×100=50a
Velocity after 10 sec. is v=0+10a
So, s2=10a×10+12a×100=150a⇒s2=3s1
Let a be constant acceleration, using s=ut+12at2
so distance coreved in first 10 seconds s1=0+12×a×100=50a
So, distance covered in next 10 seconds s2=10a×10+12a×100=150a⇒s2=3s1
The ball is thrown up and it returns in 6sec,
Time of accent = Time of decent
So, time to reach the highest point =6/2=3s
By 2nd equation of motion
s=ut+12gt2
To calculate height, consider motion of the ball from highest point to the ground
H=12gt2=12×10×32=45 m
To calculate projection velocity, consider motion of the ball from projection to return to the point of projection.
0=ut−12gt2
u=10×6/2=30 m/s
200=u×2+12a(2)2
100=u+a..........(i)
420=u(6)+12a(6)2
70=u+3a..........(ii)
from (i) and (ii) 2a=−30⇒a=−15cm/s2
∴u=100−a=100−(−15)=115cm/s
u7th=u+at=115+(−15)(7)=115−105=10cm/sec
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