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You throw a ball upwards from a starting height of 0.0m with a velocity of 4.56m/s. a) What is the vi? b) what is the velocity when the ball reaches its highest point? c) What is the acceleration when you throw the ball? d) What is the acceleration at the ball’s highest point? e) What is the velocity right as the ball reaches your hand on its way back down? (Your hand is at di=0m)

Answers

Answer 1

Answer:

Following are the solution to the given points:

Explanation:

In point a:

[tex]v_i = 4.56 \ \frac{m}{s}[/tex]

In point b:

The highest point : v=0

In point c:

[tex]a= -g \ \frac{m}{s^2} = -9.8 \frac{m}{s^2}[/tex]

In point d:  

The highest point: [tex]a= -9.8 \ \frac{m}{s}[/tex]  

In point e:

[tex]v^2=v_i^{2}+2g \Delta h \ \ \ \ \ \ Delta h=0\\\\v^2 = v_i^{2}\\\\v= 4.56 \ \ \frac{m}{s}[/tex]

Therefore, the velocity of the ball when it reaches into the hand [tex]4.56 \ \frac{m}{s}[/tex].


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