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Sagot :
h = 1/2 * g t^2
= 1/2 * 9.8 * 3^2
= 44.1 m
d = vx * t = 20 * 3 = 60 m
Ans: 44.1 m, 60 m
= 1/2 * 9.8 * 3^2
= 44.1 m
d = vx * t = 20 * 3 = 60 m
Ans: 44.1 m, 60 m
Answer:
The bridge is 44.15 m high
Explanation:
This is a case of projectile motion, the stone is projected into the air and is acted upon by force of gravitation until it ends up in the water.
Projectile motion
This can be described as a motion of an object affected by gravitational force when projected or thrown into the air and it normally has a parabolic pathway. An Example is a motion of projected missile.
The figure attached shows the man and the projectile motion of the stone.
Calculations
The equation of motion (equation 1) can be used in calculating the height of the bridge;
[tex]S = ut + \frac{1}{2} gt^{2}[/tex]........................................................... 1
where S is the distance
u is the initial velocity
g is the acceleration due to gravity
t is the time
The stone was at rest before thrown, so the initial velocity is 0 m/s;
Given u = 0 m/s
t = 3 secs
g (constant) = 9.81 [tex]m^{2}[/tex]/s
the height can be calculated by substituting the values above into equation 1
S = (0 m/s)(3 secs) + [tex]\frac{1}{2}[/tex](9.81 m/ [tex]s^{2}[/tex])[tex](3secs)^{2}[/tex]
S = 0 + 44.15 m
S = 44.15 m
Therefor the height of the bridge is 44.15 m

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