This question can be solved with the help of the equations of motion.
A) The Frisbee will remain in the air for "5.87 s".
B) The frisbee will go "29.4 m" down the range.
A)
To calculate the time, the frisbee will remain in the air, we will use the second <em><u>equation of motion</u></em>, for the vertical motion.

where,
h = height = 169.2 m
vi = initial velocity's vertical component = 0 m/s
g = acceleration due to gravity = 9.81 m/s²
t = time = ?
Therefore,

<u>t = 5.87 s</u>
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B)
Now, we will calculate the horizontal range by applying the equation for constant motion. Because the velocity in the horizontal direction will remain constant due to no air resistance
s = vt
where,
s = horizontal range = ?
v= initial velocity's horizontal component = 5 m/s
t = time = 5.87 s
Therefore,
s = (5 m/s)(5.87 s)
<u>s = 29.4 m</u>
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Learn more about <em><u>equations of motion</u></em> here:
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Answer: the distance between him and his friend is 218.39 feet
Explanation:
Given the data in the question;
FROM IMAGE A;
distance travelled in minute and a half
⇒ (60 + 30)sec × 4ft/sec = 360 fts
FROM IMAGE B
tan35°= h/x --- equ 1
and tan36° = h/(360 - x), so h = (360 - x)tan36°
we substitute value of h into euq 1
tan35° = (360 - x)tan36°/x
xtan35° = (360 - x)tan36°
0.7002x = 261.5553 - 0.7265x
0.7002x + 0.7265x = 261.5553
1.4267x = 261.5553
x = 183.32 feet
so
360 - x ⇒ ( 360 - 183.32) = 176.68
FROM IMAGE C
Let the distance between them be d
so
cos36° = 176.68 / d
dcos36° = 176.68
d = 176.68 / 0.809
d = 218.39 feet
Therefore the distance between him and his friend is 218.39 feet
a 10 kg block reaches a point with a velocity of 15 m per second and slides down a rough track my the coefficient of the kinetic energy between the two surface ab and the block iis0.52
Answer:
Yes,in fact visible 'light' is a form of radiation, which can be defined as an energy that travels in the form of electromagnetic waves. It can also be described as a flow of particle-like 'wave-packets', called photons, that travel constantly at the speed of light (about 300 000 kilometres per second).
Explanation: