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3 years ago

A volleyball player bumps a ball across a net with the velocity and angle shown below. What is the maximum height of the ball?

Physics

1 answer:

Marrrta [24]3 years ago

6 0

Answer:

D. 12.4 m

Explanation:

Given that,

The initial velocity of the ball, u = 18 m/s

The angle at which the ball is projected, θ = 60°

The maximum height of the ball is given by the formula

h = u² sin²θ/2g m

Where,

g - acceleration due to gravity. (9.8 m/s)

Substituting the values in the above equation

h = 18² · sin²60 / 2 x 9.8

= 18² x 0.75 / 2 x 9.8

= 12.4 m

Hence, the maximum height of the ball attained, h = 12.4 m

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Liono4ka [1.6K]

Average velocity is defined as the ratio of displacement and time

Here we know that the total distance is the sum of all given distances but we need the displacement in order to find the velocity

So here displacement is straight line distance or direct distance from her home to her daughter home

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3 years ago

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Scorpion4ik [409]

Answer:

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Explanation:

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3 years ago

Two cello strings, with the same tension and length, are played simultaneously. Their fundamental frequencies produce audible be

qwelly [4]

Explanation:

Let f₁ is the fundamental frequency, $f_1=8\ Hz$

Lower pitch frequency, $f_2=220\ Hz$

Fundamental frequency is, $f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu_1}}$ .....(1)

Lower frequency is, $f_2=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu_2}}$ ..............(2)

Dividing equation (1) and (2) as :

$\dfrac{f_1}{f_2}=\sqrt{\dfrac{\mu_2}{\mu_1}}$

$\dfrac{\mu_2}{\mu_1}=(\dfrac{f_1}{f_2})^2$

$\dfrac{\mu_2}{\mu_1}=(\dfrac{8}{220})^2$

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So, the ratio of linear mass density μ of the string with the higher pitch to that of the string with the lower pitch is 0.00132. Hence, this is the required solution.

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Dahasolnce [82]

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