Of the cliff?
Projectile motion
In the problem we are asked to find a height of certain cliff when a motorcycle stunt driver zoom out horizontally at the end the cliff at an initial velocity. So we will use one of the kinematics equation for projectile motion,
y
=
v
o
y
t
+
1
2
g
t
where
v
o
y
is just equal to zero since we can assume that the driver zooms out horizontally,
g
=
9.8
m
/
s
2
and
t
is time after
The highest elevation reached by the ball in its trajectory is 16.4 m.
To find the answer, we need to know about the maximum height reached in a projectile.
What's the mathematical expression of the maximum height reached in a projectile motion?
- The maximum height= U²× sin²(θ)/g
- U= initial velocity, θ= angle of projectile with horizontal and g= acceleration due to gravity
What's the maximum height reached by a block that is thrown with an initial velocity of 30.0 m/s at an angle of 25° above the horizontal?
- Here, U = 30.0 m/s and θ= 25°
- Maximum height= 30²× sin²(25)/9.8
= 16.4m
Thus, we can conclude that the highest elevation reached by the ball in its trajectory is 16.4 m.
Learn more about the projectile motion here:
brainly.com/question/24216590
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2.c
3.b
1.a
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Answer:
Ongoing effects include rising sea levels due to thermal expansion and melting of glaciers and ice sheets, and warming of the ocean surface, leading to increased temperature stratification. Other possible effects include large-scale changes in ocean circulation
Explanation:
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