A curve ball is a type of pitch in which the baseball spins on its axis as it heads for home plate. If a curve ball is thrown at
1 answer:
Answer:
19.35 revolutions.
Explanation:
Let the home plate is 18.3 m (60 ft) from the pitching mound and that the baseball travels at a constant velocity.
Given that,
Velocity of ball = v = 34.5 m/s
Spin rate of ball = 26 revolutions/s
Distance between home plate and pitching mound = s = 18.3 m
We need to find the number of revolutions it complete before reaching home plate. It can be calculated as follows:
So, there are approx 19.35 revolutions.
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Answer:
t = 16.5 s
Explanation:
First we apply first equation of motion to the accelerated motion of the rocket:
where,
vf₁ = final speed of rocket during accelerated motion = ?
vi₁ = initial speed of rocket during accelerated motion = 0 m/s
a = acceleration of rocket during accelerated motion = 30 m/s²
t₁ = time taken during accelerated motion = 4 s
Therefore,
Now, we analyze the motion rocket when engine turns off. So, the rocket is now in free fall motion. Applying 1st equation of motion:
where,
vf₂ = final speed of rocket after engine is off = 0 m/s
vi₂ = initial speed of rocket after engine is off = Vf₁ = 120 m/s
g = acceleration of rocket after engine is off = - 9.8 m/s² (negative sign for upward motion)
t₂ = time taken after engine is off = ?
Therefore,
So, the time taken from the firing position till the stopping position is:
<u>t = 16.5 s</u>