Answer :
here,
u=initial speed = 0.
t = time = 0.5 sec.
a= acceleration = g = 9.8 m/sec^2
findout distance:-
s=ut+[tex] \frac{1}{2} at^{2} [/tex]
s=0*t +[tex] \frac{1}{2}*9.8* \frac{1}{2}* \frac{1}{2}= \frac{9.8}{8}=1.225m[/tex]
u=initial speed = 0.
t = time = 0.5 sec.
a= acceleration = g = 9.8 m/sec^2
findout distance:-
s=ut+[tex] \frac{1}{2} at^{2} [/tex]
s=0*t +[tex] \frac{1}{2}*9.8* \frac{1}{2}* \frac{1}{2}= \frac{9.8}{8}=1.225m[/tex]
using equation s = ut +
s = distance
u=initial speed = 0.
t = time = 0.5 sec.
a= acceleration = g = 9.8 m/s²
⇒ s = 0 × 0.5 + 1/2 × 9.8 × 0.5 ×0.5
⇒ s = 1.2 mtrs.
s = distance
u=initial speed = 0.
t = time = 0.5 sec.
a= acceleration = g = 9.8 m/s²
⇒ s = 0 × 0.5 + 1/2 × 9.8 × 0.5 ×0.5
⇒ s = 1.2 mtrs.