Answer:
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Answer:
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Answer:
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Explanation:
Hi there!
The equations of height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity of the ball at time t.
Placing the origin at the throwing point, y0 = 0.
Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.
v = v0 + g · t
6.00 m/s = 12.0 m/s -9.81 m/s² · t
(6.00 - 12.0)m/s / -9.81 m/s² = t
t = 0.612 s
Now, let´s calculate the height of the baseball at that time:
y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)
y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²
y = 5.51 m
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Have a nice day!
Answer:
Explanation:
Given that:
the initial angular velocity 
angular acceleration
= 4.44 rad/s²
Using the formula:

Making t the subject of the formula:

where;

∴

t = 0.345 s
b)
Using the formula:

here;
= angular displacement
∴



Recall that:
2π rad = 1 revolution
Then;
0.264 rad = (x) revolution

x = 0.042 revolutions
c)
Here; force = 270 N
radius = 1.20 m
The torque = F * r

However;
From the moment of inertia;

given that;
I = 84.4 kg.m²

For re-tardation; 
Using the equation



t = 0.398s
The required time it takes= 0.398s