A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height of 60 m at the time of dropping the ball, how long will the ball take in reaching the ground ?Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics
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Answered by
154
Solution :
Initially ball is going upwards
Initial velocity=u= -7m/s
S=60m
a=g=10m/s2
From Second equation of motion
s=ut+1/2at²
60= - 7t +(1/2 )10 t²
60=-7t +5t²
5t² -7t -60=0
t=-7±√49-4x5(-60)/2x5
t=-7±√49+1200/10
t=-7±35.3/10
t=7+35.3/10 =4.2 sec [ as t cannot be negative]
∴ The ball will take 4.2 sec to reach the ground.
Initially ball is going upwards
Initial velocity=u= -7m/s
S=60m
a=g=10m/s2
From Second equation of motion
s=ut+1/2at²
60= - 7t +(1/2 )10 t²
60=-7t +5t²
5t² -7t -60=0
t=-7±√49-4x5(-60)/2x5
t=-7±√49+1200/10
t=-7±35.3/10
t=7+35.3/10 =4.2 sec [ as t cannot be negative]
∴ The ball will take 4.2 sec to reach the ground.
Answered by
55
HEY!!
______________________________
✔s = ut + (1/2)at2
✔s = 60 m
✔u = -7 m/s
✔a = 10 m/s^2
✔As per the formula 60 = -7t + 5t^2
✔-5t^2 + 7t = 60
✔By solving the above quadratic equation further, we will get
▶▶t = 4.23 secs
______________________________
✔s = ut + (1/2)at2
✔s = 60 m
✔u = -7 m/s
✔a = 10 m/s^2
✔As per the formula 60 = -7t + 5t^2
✔-5t^2 + 7t = 60
✔By solving the above quadratic equation further, we will get
▶▶t = 4.23 secs
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