A man is standing at the top of 105 meter high tower. He throws a ball vertically upwards, with a velocity of 20m/s. After how many seconds will the ball pass him going downward? How long after its release will the ball reach the ground?
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google pe kar sakta hai
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u=20m/s
v=om/s
let g=10m/s2
v=u+at
o=20+(-10)*t(a=g)
0 = 20-10t
10t=20
t=20/10=2sec
now, v^{2}=u^{2}+2-gh
o =20*20+2*-10*h
0 =400-20h
20h=400
h=400/20=20m
hence total height ground to max height=105+20=125m
v=om/s
let g=10m/s2
v=u+at
o=20+(-10)*t(a=g)
0 = 20-10t
10t=20
t=20/10=2sec
now, v^{2}=u^{2}+2-gh
o =20*20+2*-10*h
0 =400-20h
20h=400
h=400/20=20m
hence total height ground to max height=105+20=125m
anishranjan:
it may help u
sry yrr i"m forgot to gis going upward so its sign=-g(-10m/s spuare
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