"A ball launched at a speed of 20.0 m/s at angle of 40.0°C above the horizontal."
What is the maximum height reached by the ball?
Please I need the solution and answer also, thank you very much!
What is the maximum height reached by the ball?
Please I need the solution and answer also, thank you very much!
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Answer:
Example 1
A cannon ball is fired with an initial velocity of 100.0 m/s at an angle of 45° above the horizontal. What maximum height will it reach and how far will it fly horizontally?
The first step in the analysis of this motion is to resolve the initial velocity into its vertical and horizontal components.
vi−up=(100.0 m/s)(sin45∘)=(100.0 m/s)(0.707)=70.7 m/s
vi−horizontal=(100.0 m/s)(cos45∘)=(100.0 m/s)(0.707)=70.7 m/s
We will deal with the vertical motion first. The vertical motion is symmetrical. As the object rises to its highest point and then falls back down, it will travel the same distance in each direction, and take the same amount of time. This is often hard to accept, but the amount of time the object takes to come to a stop at its highest point is the same amount of time it takes to return to where it was launched from.
Explanation:
it's just an example