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suppose a ball is thrown vertically upward. Eight seconds later it returns to its point of release. What is the initial velocity of the ball?

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AL2006
The ball took half of the total time ... 4 seconds ... to reach its highest
point, where it began to fall back down to the point of release.

At its highest point, its velocity changed from upward to downward. 
At that instant, its velocity was zero.

The acceleration of gravity is 9.8 m/s².  That means that an object that's
acted on only by gravity gains 9.8 m/s of downward speed every second. 

-- If the object is falling downward, it moves 9.8 m/s faster every second.

-- If the object is tossed upward, it moves 9.8 m/s slower every second.

The ball took 4 seconds to lose all of its upward speed.  So it must have
been thrown upward at  (4 x 9.8 m/s)  =  39.2 m/s .

(That's about  87.7 mph straight up.  Somebody had an amazing pitching arm.)
Eduard22sly

The initial velocity of the ball thrown vertically upward is 40 m/s.

How to determine the time

We'll begin by calculating the the time taken for ball to reach maximum height. This can be obtained as follow:

  • Time of flight (T) = 8 s
  • Time to reach maximum height (t) =?

T = 2t

8 = 2t

Divide both side by 2

t = 8 / 2

t = 4 s

How to determine the initial velocity

  • Time to reach maximum height (t) = 4 s
  • Final velocity (v) = 0 m/s (at maximum height)
  • Acceleration due to gravity (g) = 10 m/s²
  • Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (10 × 4)

0 = u – 40

Collect like terms

u = 0 + 40

u = 40 m/s

Learn more about motion under gravity:

https://brainly.com/question/22719691

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