The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin
θ.
The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.
If the <u>air resistance is ignored</u>, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.
Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
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The answer is D, the amount of energy stays the same.
Answer:
(a) 41.75m/s
(b) 4.26s
Explanation:
Let:
Distance, D = 89m
Gravity, 
= 9.8 m/
Initial Velocity, 
= 0m/s
Final Velocity, 
= ?
Time Taken, 
= ?
With the distance formula, which is
D = 
+ 
and by substituting what we already know, we have:
89 = 
×9.8×
With the equation above, we can solve for :
Now that we have solved , we can use the following velocity formula to solve for :
, where 
is also equals to , so we have
By substituting 
, 
, and 
,
We have:
Part of the cell that stores food is called vacuole
<span>The use of the word on instead of the word in when referring to the angular distance between celestial objects comes about because all of the objects appear to be on the celestial sphere and at an indeterminable distance. While we know that objects are at different distances in the sky, their distance from Earth is irrelevant in determining the angular distance between the two objects as viewed from Earth.</span>
