Answer:
when the ball is at its highest in the air.
Explanation:
I don't know for sure, but when the ball is in the air it has potential energy to fall(or something like that).
Answer:
(a) a = - 201.8 m/s²
(b) s = 197.77 m
Explanation:
(a)
The acceleration can be found by using 1st equation of motion:
Vf = Vi + at
a = (Vf - Vi)/t
where,
a = acceleration = ?
Vf = Final Velocity = 0 m/s (Since it is finally brought to rest)
Vi = Initial Velocity = (632 mi/h)(1609.34 m/ 1 mi)(1 h/ 3600 s) = 282.53 m/s
t = time = 1.4 s
Therefore,
a = (0 m/s - 282.53 m/s)/1.4 s
<u>a = - 201.8 m/s²</u>
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(b)
For the distance traveled, we can use 2nd equation of motion:
s = Vi t + (0.5)at²
where,
s = distance traveled = ?
Therefore,
s = (282.53 m/s)(1.4 s) + (0.5)(- 201.8 m/s²)(1.4 s)²
s = 395.54 m - 197.77 m
<u>s = 197.77 m</u>
Answer:
i hope the answear is D becuase went over this long time ago when i was like you
Explanation:
Answer: The minimum acceleration for the air plane is 2.269m/s2.
Explanation: To solve such problem the equation of motion are applicable.
The initial velocity is 0 since the airplane was initially standing. We are going to use this equation
V^2=U^2+2as
33^2=0+2a (240)
a= 2.269m/s2
