I have 2 controversy question
1. A girl standing on a stationary lift throws a ball upwards with initial speed 50 m/s the time taken by the ball to return to her hands is?
2 .a player throws a ball upwards with an initial speed 30m/s how long does the ball take to return to the players hand?
Why can't we I use 1st law of motion for the first case(v=u+at)
rohirestle:
Pls guys give me a related one
Answers
Answered by
1
1. Simply apply 2nd equation as,
Displacement = 0
Initial velocity (u) = 50 m/s
a = - g = - 10 m/s^2
s = ut + 1/2 at^2
0 = 50(t) + 1/2 (-10) t^2
0 = 5t (10 - t)
Total time taken by ball to return to her hands = 10 sec
U can also apply 1 equation but when u apply 1 equation the time u will get will be the time taken by ball to reach maximum height.... So in order to calculate total time... U have to multiply calculated time by 2...
Similarly u can solve 2 question
Hope it helps... ☺
Displacement = 0
Initial velocity (u) = 50 m/s
a = - g = - 10 m/s^2
s = ut + 1/2 at^2
0 = 50(t) + 1/2 (-10) t^2
0 = 5t (10 - t)
Total time taken by ball to return to her hands = 10 sec
U can also apply 1 equation but when u apply 1 equation the time u will get will be the time taken by ball to reach maximum height.... So in order to calculate total time... U have to multiply calculated time by 2...
Similarly u can solve 2 question
Hope it helps... ☺
i think answer of 2 question should be 6 sec
HEY BRILLIANT
Thnx
i too got but i couldnt understant that
what problem r u facing with??
the thing i use 1st equation i get only the time to reach maximum i need clarity yar pls
when body is moving upward then v = 0 so when u put this value in 1 equation u will get the time taken by ball upto the point when its velocity will be 0
oh my goodness i got it
but when body is moving downward it has some final velocity so that's why 2 equation is more appropriate
OKAYYY
Similar questions