a ball is dropped from height H. if it takes 0.2seconds to cover last 6metre before touching the ground find H.
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For last 6 m distance travelled
s = 6 m,
u = ?
t = 0.2 sec,
a = g = 9.8 m/s2
S = ut + ½ at2 ⇒
6 = u (0.2) + 4.9 x 0.04 ⇒
u = 5.8/0.2 = 29 m/s.
For distance x, u = 0,
v = 29 m/s,
a = g = 9.8 m/s2 S
= (V^2-u^2)/2a
= (〖29〗^2 - 0^2 )/(2 x 9.8 )
= 42.05 m
Total distance = 42.05 + 6 = 48.05 = 48 m
For last 6 m distance travelled
s = 6 m,
u = ?
t = 0.2 sec,
a = g = 9.8 m/s2
S = ut + ½ at2 ⇒
6 = u (0.2) + 4.9 x 0.04 ⇒
u = 5.8/0.2 = 29 m/s.
For distance x, u = 0,
v = 29 m/s,
a = g = 9.8 m/s2 S
= (V^2-u^2)/2a
= (〖29〗^2 - 0^2 )/(2 x 9.8 )
= 42.05 m
Total distance = 42.05 + 6 = 48.05 = 48 m
avadacrucio:
thank you so much..
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