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3 years ago

9

A ball thrown vertically upward is caught by the thrower after 1.76 s. find the initial velocity of the ball. the acceleration o

f gravity is 9.8 m/

1 answer:

konstantin123 [22]3 years ago

7 0

Answer;

= 8.624 m/s

Explanation;

Time of flight= 4s. (0.88 s to reach highest point and 0.88 s to come back from highest point.)

Neglecting air resistance, the time of ascend is equal to time of descend for projectile.

It means the ball is at the highest point at time 0.88 s.

Using the equation;

V = Vo + at

Where, V is the final velocity which is Zero, Vo is the initial velocity ,g is the gravitational acceleration, 9.8 m/s², t is the time.

V = Vo - gt ; g is negative because it is against the gravity .

Vo = gt

= 9.8 m/s² × 0.88 s

= 8.624 m/s

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$-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)$

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$t_{1}=g*1.06/(2.06*g-2v_{o})$

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$g*1.06/(2.06*g-2v_{o})>0$

$2.06*g-2v_{o}>0$

$v_{o}$

$v_{max}$

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