Answer:
The velocity of the rock is 47.24 m/s
The displacement of the rock is 110.48 m
Explanation:
The initial velocity u = 8.0 m/s
time t = 4.0 s
Using
[tex]v = u + gt[/tex]
where v is the final velocity of the rock after 4.0 s
u is the initial speed = 8 m/s
g is acceleration due to gravity = 9.81 m/s
t is the time = 4.0 s
substituting, we have
[tex]v = 8 + (9.81*4)[/tex]
[tex]v = 8 + 39.24[/tex] = 47.24 m/s
Also, using
[tex]s = ut + \frac{1}{2} gt^{2}[/tex]
where
s is the distance that the stone fell
t is the time = 4.0 s
g is the acceleration due to gravity = 9.81 m/s^2
substituting values, we have
[tex]s = (8 * 4) + (\frac{1}{2}*9.81*4^{2})[/tex]
s = 32 + 78.48
s = 110.48 m
Given:
By using the relation, we get
→ [tex]v = u+gt[/tex]
By substituting the values, we get
[tex]= 8+(9.8\times 4)[/tex]
[tex]= 8+39.24[/tex]
[tex]= 47.24 \ m/s[/tex]
hence,
→ [tex]s = ut+\frac{1}{2} gt^2[/tex]
[tex]= (8\times 4)+(\frac{1}{2}\times 9.81\times 4^2 )[/tex]
[tex]= 32+78.48[/tex]
[tex]= 110.48 \ m[/tex]
Thus the above answer is appropriate.
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