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3 years ago

A ball is thrown with an initial upward velocity of 5 m/s.

Physics

1 answer:

miv72 [106K]3 years ago

6 0

Answer:

A ball is thrown at an initial height of 5 feet with an initial upward velocity at 29 ft/s. lets assume that balls height h (in feet) after t seconds is give by:

Explanation:

h= 5 + 29t -16t^2

a time when the ball's height will be 17 ft

17 = 5 + 29t -16t2

0 = -17 + 5 + 29t -16t2

0 = -12 + 29t - 16t2

Using the quadratic equation:

t = (-29±√(292-(4*(-16)*(-12))))÷2(-16)

= (-29±√(841 - 768))÷(-32)

= (-29±√(73))÷(-32)

= (-29 + 8.544)÷(-32) or (-29 - 8.544)÷(-32)

= (-20.456)÷(-32) or -37.544÷(-32)

= 0.64 or 1.17

So, the ball is at a height of 17 ft twice: once on the way up after 0.64 seconds and once on the way back down after 1.17 seconds.

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Explanation:

From the we are told that

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Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

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