Answer:
400ft. 32ft/s -32ft/s
Explanation:
In reality the gravitational acceleration is 9.81 so the quadratic coefficient of the function should be 9.81/2
Anyway for the sake of assumtion let us takes=160t-16t^2
ds/dt=160-32t=0
t=160/32= 5 seconds.
s=160*160/32-16*(160/32)^2= 400 mts
s=384 mts
160t-16t^2=384
i.e
16t^2-160t+384=0
t^2-10t+24=0
(t-6)(t-4)=0
t=[4,6]
we have to take t=4 because it is all the up i.e <5
velocity =v=ds/dt=160-32t
v=160-32*4=32 ft/sec still going up
for all the way down take t=6 whuch is >5
v=160-6*32=-32 ft/sec (falling down!!!)
To answer the two questions, we need to know two important equations involving centripetal movement:
v = ωr (ω represents angular velocity <u>in radians</u>)
a = 
Let's apply the first equation to question a:
v = ωr
v = ((1800*2π) / 60) * 0.26
Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.
Back to the equation:
v = ((1800*2π)/60) * 0.26
v = (1800*2(3.14)/60) * 0.26
v = (11304/60) * 0.26
v = 188.4 * 0.26
v = 48.984
v = 49 (m/s)
Now that we know the linear velocity, we can find the centripetal acceleration:
a =
a = 
a = 9234.6 (m/
)
Wow! That's fast!
<u>We now have our answers for a and b:</u>
a. 49 (m/s)
b. 9.2 * 
(m/)
If you have any questions on how I got to these answers, just ask!
- breezyツ
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
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The boat traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.
Explanation:
Velocity is a physical quantity that describes the rate of change of displacement with time.
Velocity = 
The quantity differs from speed in that it has both magnitude and direction.
From the options given above:
Displacement: The boat traveled in the north direction from the dock to a 200m mark.
Time taken: approximately less than 5 minutes was the duration of traveling.
This describes the boat's velocity accurately.
Learn more:
Velocity
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