A man throws a ball vertically upward and it rises through 20m & returns to his hands. What was the initial velocity of the ball and for how much time(T) it remains in the air [g=10m/s^2]
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the time taken for it to fall down is given by
s=ut + 1/2at^2
s=0×the + 0.5×10×t^2
20=5t^2
4=t^2
t=2s
now twice that time is the total time taken ( time of ascent = time of descent)
therefore total time = T = 4s.
final velocity = 0 ( for ascent)
v= U + at
0=U + (-10×2)
u= 20m/s.
s=ut + 1/2at^2
s=0×the + 0.5×10×t^2
20=5t^2
4=t^2
t=2s
now twice that time is the total time taken ( time of ascent = time of descent)
therefore total time = T = 4s.
final velocity = 0 ( for ascent)
v= U + at
0=U + (-10×2)
u= 20m/s.
ajjubhai124:
ha bhai maine bhi yahi ans nikala pr ans uska u=20m/s, T=4s
diya option me
Answered by
1
h=20 g=10 u=? u=√2gh =√2×10×20 =√400 u =20 Now time- t=√2gh =√2×20\10 = √4 t=2
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