Respuesta :
Answer:
Ball will move 92.8125 meter along the cliff in 7.5 seconds.
Explanation:
We have equation of motion , [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case initial velocity = 0 m/s, acceleration = 3.3 [tex]m/s^2[/tex], we need to calculate displacement when time = 7.5 seconds.
Substituting
[tex]s=0*7.5+\frac{1}{2} *3.3*7.5^2\\ \\ =92.8125 meter[/tex]
So ball will move 92.8125 meter along the cliff in 7.5 seconds.
The ball will move through a distance of 84.375 m
From the question,
The ball is initially at rest before rolling down the hill
This means the initial velocity, u, of the ball is 0 m/s
To calculate how far the ball move,
From one of the equations of kinematic for linear motion
We have that
[tex]s = ut + \frac{1}{2}at^{2}[/tex]
Where,
s is the distance
u is the initial velocity
a is the acceleration
and t is the time
From the question
u = 0 m/s
a = 3.3 m/s²
t = 7.5 s
Putting these values into the equation
[tex]s = ut + \frac{1}{2}at^{2}[/tex]
We get
[tex]s = 0(7.5) + \frac{1}{2} (3.3)(7.5)^{2}[/tex]
[tex]s = 0+ \frac{1}{2} \times 3.3 \times 7.5^{2}[/tex]
[tex]s = \frac{1}{2} \times 3.3 \times 56.25[/tex]
[tex]s = \frac{168.75}{2}[/tex]
s = 84.375 m
Hence, the ball will move through a distance of 84.375 m
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