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A ball is thrown upward with an initial vertical velocity of v0 to a maximum height of hmax and falls back toward earth. on the way down, it is caught at the same height at which it was thrown upward. b b b b b b b b b bbb b b b b b b b b v0 hmax if the total time the ball remains in the air is t, find its speed when caught. the acceleration of gravity is g. neglect air resistance.

Respuesta :

Answer: [tex]v_f=g\frac{t}{2}[/tex]

The ball was thrown at the speed of [tex]v_o[/tex].

Maximum height achieved is [tex]h_{max}[/tex]

Time of flight is t.

Now, the time the ball takes to achieve maximum height = the time taken by ball to fall back = [tex]\frac{t}{2}[/tex]

let us just consider the second half of the flight. At [tex]h_{max}[/tex], the velocity would be zero. let us consider as the initial velocity for the second half of the flight i.e. [tex]v_i=0[/tex]

Using the equation of motion:

[tex]v_f=v_i+at[/tex]

where,  [tex]v_f[/tex] is the final velocity, a is the acceleration, t is the time taken.

Because the ball would fall under gravity, hence a=g and time of flight would be t/2

[tex]\Rightarrow v_f=0+g\frac {t}{2}[/tex]

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