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3 years ago

7

A soccer ball is kicked from the ground at an angle of θ = 42 degrees with respect to the horizontal. The ball is in the air for

a time tm = 1.3 s before it lands back on the ground.Numerically, what is the total horizontal distance, dm in meters, traveled by the ball in the time, tm?

1 answer:

umka21 [38]3 years ago

3 0

Answer:

Explanation:

Given

Launch angle $\theta=42^{\circ}$

time period of flight $T=1.3\ s$

Time period of Projectile $T=\frac{2u\sin \theta }{g}$

Where u=initial velocity

g=acceleration due to gravity

$1.3=\frac{2\times u\times \sin (42)}{9.8}$

$u=9.51\ m/s$

Range of Projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{9.51^2\times \sin (84)}{9.8}$

$R=\frac{90.44\times 0.9945}{9.8}=9.17\ m$

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A wet bar of soap slides down a ramp 9.2 m long inclined at 8.0∘ .

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Answer:

The time taken by the bar to reach the bottom t=4.886s

Given:

Displacement of the bar S=9.2m

Angle of inclination $\theta=8.0^{\circ}$

Coefficient of friction factor $\mu k=0.056$

To find:

How long it takes to reach the bottom ‘t’

<u>Step by Step Explanation:</u>

Solution:

We know that the formula for weight of the soap bar is given as

$F_{g}=m g \sin \theta$

The frictional force acting on this soap bar is determined by

$F_{f}=\mu m g \cos \theta$

To determine the constant acceleration of the bar, we derive as

$F=m a$

Here $F=F_{g}-F_{f}$ and thus

$F_{g}-F_{f}=m a$ $m g \sin \theta-\mu m g \cos \theta=m a$

$a=g \sin \theta-\mu g \cos \theta$

Where $F_{g}$ =Force imparted due to weight

$F_{f}$ =Frictional Force

m=Mass of the bar

g=Acceleration due to gravity

a=Acceleration of the bar

$\sin \theta$ and $\cos \theta$ are the angles involved in the system

If the bar starts from the rest

Equations of motion involved in calculating the displacement of the bar is given as

$s=\frac{1}{2} a t^{2}$ , From this

$a t^{2}=2 s$

$t^{2}=\frac{2 s}{a}$

$t=\sqrt{\frac{2 s}{a}}$

Where s= displacement or length moved by the bar

a=Acceleration of the bar

t=Time taken to reach bottom

Substitute all the known values in the above equation we get

$t=\sqrt{\frac{2 \times 9.2}{a}}$ and we know that

$a=g \sin \theta-\mu g \cos \theta$

$=9.8 \times \sin 8.0-0.056 \times 9.8 \times \cos 8.0$

$=1.364-0.543$

$a=0.821$

$t=\sqrt{\frac{2 \times 9.2}{0.821}}$

$t=\sqrt{\frac{19.6}{0.821}}$

$t=\sqrt{23.87332}$

t=4.886s

Result:

Thus the time taken by the bar to reach the bottom is t=4.886s

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