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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When two tectonic plates collide and form a converging plate boundry, normally one of the plates will slide underneath the other and that is when Subduction occurs.
According to Ohm's law for a portion of the circuit we have:
U=RI=>I=U/R=24/3=8 A
The correct answer is B
Answer:
2.28
Explanation:
From mirror formula,
1/f = 1/u+1/v .......... Equation 1
Where f = focal length of the mirror, v = image distance, u = object distance.
Note: The focal length mirror is positive.
make v the subject of the equation,
v = fu/(u-f)............ Equation 2
Given: f = 2.5 cm, u = 1.4 cm
Substitute into equation 2
v = 2.5(1.4)/(1.4-2.5)
v = 3.5/-1.1
v = -3.2 cm.
Note: v is negative because it is a virtual image.
But,
Magnification = image distance/object distance
M = v/u
Where M = magnification.
Given: v = 3.2 cm, u = 1.4 cm
M = 3.2/1.4
M = 2.28.
Thus the magnification of the tooth = 2.28.
