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3 years ago

A ball player catches a ball 3.13 s after throwing it vertically upward. a)With what speed did he throw it?

Physics

1 answer:

Anna35 [415]3 years ago

4 0

Answer:

a) v₀ = 15.3 m/s

b) h = 11.9 m

Explanation:

Ball Kinematics

We apply the free fall formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*y Formula (2)

y= v₀t+ (1/2)*a*t² Formula (3)

y:displacement in meters (m)

v₀: initial speed in m/s

f: final speed in m/s

a: acceleration in m/s²

Data

t= 3.13 s

a=g= -9.8 m/s² = acceleration due to gravity

Problem development

a)With what speed did he throw it

t= 3.13 s : total ball time going up and down
The time the ball rises to its maximum height is half the total time
At maximum height vf = 0

We apply the formula (1) for calculate initial speed (v₀)

vf= v₀+at

0= v₀+(-9.8)*(3.13/2)

v₀= 15.3 m/s

b)What height did it reach?

We apply formula (2) or formula (3)

vf²=v₀²+2*a*h formula (2)

0 = v₀² +2*(-9.8)*h

19.6*h = v₀²

h=v₀²/19.6

h= (15.3)²/(19.6 )

h= 11.9m

y= v₀t+ (1/2)*a*t² Formula (3)

h= (15.3)(3.13/2)+(1/2)(-9.8)(3.13/2)²= 11.9m

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4 years ago

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45 m diamete

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Answer:

In m/s^2:

a=11.3778 m/s^2

In units of g:

a=1.161 g

Explanation:

Since the racing greyhounds are capable of rounding corners at very high speed so we are going use the following formula of acceleration for circular paths.

$a=\frac{v^2}{r}$

where:

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3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

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Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

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Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

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