a) The final speed of the ball is 49 m/s
b) The height of the bridge is 122.5 m
Explanation:
a)
The motion of the ball is a free fall motion, so it is a uniformly accelerated motion with constant acceleration [tex]g=9.8 m/s^2[/tex] (acceleration of gravity) towards the ground. Therefore, we can use the following suvat equation:
[tex]v=u+at[/tex]
where, taking downward as positive direction:
v is the final velocity (at time t)
u is the initial velocity
a is the acceleration
For the ball in this problem, we have
u = 0 (it is dropped from rest)
[tex]a=g=9.8 m/s^2[/tex]
Substituting t = 5 s, we find the final velocity and speed of the ball:
[tex]v=0+(9.8)(5)=49 m/s[/tex]
b)
The height of the bridge can be found by using the other suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s is the vertical displacement of the ball (= the height of the bridge)
u is the initial velocity
a is the acceleration
t is the time
For the ball in this problem, we have
u = 0
[tex]a=g=9.8 m/s^2[/tex]
t = 5 s (time of the fall)
Solving for s, we find the height of the bridge:
[tex]s=0+\frac{1}{2}(9.8)(5)^2=122.5 m[/tex]
Learn more about free fall motion:
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