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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Answer:
If an object has a fast velocity, the dots on a ticker tape diagram will be far apart.
Answer:
40N in either direction is the answer
Answer:
16.7 s
Explanation:
T= <u>Vf - Vo</u> a= <u>F</u>
a m
4,500 / 3000 = 1.5 (a)
30 - 5 / 1.5(a) = 16.7 s