Question

A ball is thrown from ground at an angle theta with horizontal and with an initial speed u no for the resulting projectile motion the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is given after hitting the ground by the ball rebounds at the same angle theta with reduced speed of mu not buy it

Answer 1

Answer:

the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is $v cos\theta$

Explanation:

As we know that average velocity is defined as the ratio of displacement and total time

so here we know that total displacement of the ball is equal to the range of the ball

so we will have

$d = \frac{2v^2 sin\theta cos\theta}{g}$

now total time interval of the motion is given as

$T = \frac{2vsin\theta}{g}$

now the velocity is given as

$v_{avg} = \frac{d}{T}$

$v_{avg} = \frac{\frac{2v^2 sin\theta cos\theta}{g}}{\frac{2vsin\theta}{g}}$

$v_{avg} = v cos\theta$

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Topic : Projectile Motion

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