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3 years ago

7

A ball thrown straight upwards with an initial velocity of 15 m/s. Calculate how long the ball will remain in the air and the ma

ximum height the ball reaches?

1 answer:

GenaCL600 [577]3 years ago

3 0

Answer:

The ball remain in air for 3.06 seconds.

Maximum height reached = 22.94 m

Explanation:

We have equation of motion v = u + at

At maximum height, final velocity, v =0 m/s

Initial velocity = 15 m/s

acceleration = -9.81 m/s²

Substituting

0 = 15 -9.81 t

t = 1.53 s

Time of flight = 2 x 1.53 = 3.06 s

The ball remain in air for 3.06 seconds.

We also have equation of motion v² = u² + 2as

At maximum height, final velocity, v =0 m/s

Initial velocity = 15 m/s

acceleration = -9.81 m/s²

Substituting

0² = 15² - 2 x 9.81 x s

s = 22.94 m

Maximum height reached = 22.94 m

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Answer:

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Explanation:

<u>Motion With Constant Acceleration </u>

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

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The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Solving [1] for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

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The distance is now calculated with [2]:

$\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}$

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