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denis23 [38]
3 years ago
5

A ball is throw at an angle of 30 degrees off the horizontal, with an initial velocity of 28 m/s. what is the maximum height the

ball will reach?​
Physics
1 answer:
aksik [14]3 years ago
8 0

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Angle \ of \ projection = 30^{\circ} }

\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ projectile = 28 \: m/s^{-1} }

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Height_{\:(max)}\: reached\: by \:the \:projectile }

\\

{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\

☯ <u>As</u> <u>we</u> <u>know</u> <u>that</u>,

\\

\dashrightarrow\:\: \sf{ H = \dfrac{u^2\;sin^2\theta}{2\;g} }

\\

\dashrightarrow\:\: \sf{H = \dfrac{(28)^2\;sin^2 30^{\circ}}{2\;(9.8)}  }

\\

\dashrightarrow\:\: \sf{H = \dfrac{784 \times \;sin^230^{\circ}}{19.6}  }

\\

\dashrightarrow\:\: \sf{  H = \dfrac{784}{19.6}\times sin^2 30^{\circ}}

\\

\dashrightarrow\:\: \sf{  H = \dfrac{784}{19.6}\times \bigg(\dfrac{1}{2}\bigg)^2 }

\\

\dashrightarrow\:\: \sf{  H = \dfrac{784}{19.6}\times \dfrac{1}{4} }

\\

\dashrightarrow\:\: {\boxed{\sf{H=10\:m  }}}

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For a fixed amount of gas at constant temperature (Boyle's law) : P₁V₁ = P₂V₂

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3 years ago
At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a window 10 m below the
dsp73

Answer:

Explanation:

Given two objects are dropped simultaneously

Object A is 10 m higher than object B therefore

Distance covered by object A is given by

y_a(t) is given by

y=ut+\frac{1}{2}at^2

where y=displacement

u=initial velocity

a=acceleration

t=time

y_a(t)=0+0.5gt^2--1

for object B

y_b(t)=0+0.5gt^2--2

Subtract 1 and 2 we get

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A football is kicked from ground level with an initial velocity of 23.6 m/s at angle of 57.5° above the horizontal. How long, in
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A 26.0 kg beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal. If
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Answer:

Force exerted by the hinge on the beam = 109.24N

Explanation:

Weight = mg = 26 x 9.81 = 255.06 N

Vertical component = T sin θ

Horizontal component = Tcos θ

Now, there are 3 vertical forces acting on the beam. These are;

- The downward force which is the weight of the beam.

- The vertical components of the tension in the cable.

-The force that hinge exerts on the beam are the upward forces.

Hence, for the beam to remain horizontal, the sum of the upward forces must be equal to the weight of the beam.

For us to determine the vertical component of the tension in the cable, we will do a torque problem. Let the pivot point be at the hinge. Let’s assume that the length of the beam is L. The vertical component of the tension in the cable will produce clockwise torque while the weight of the beam will produce counter clockwise torque.

Tbus;

Clockwise torque = TL sin 61

Since the center of mass of beam is at the middle of the beam, the distance from the hinge to the weight of the beam is L/2.

Counter clockwise torque = WL/2

Thus;

TL sin 61 = WL/2

L will cancel out.

T sin 61 = 255.06/2

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T = 127.53/0.8746 = 145.82 N

Now, the equation to determine the vertical component of the force that the hinge exerts on the beam is given as;

T + F = W

Thus;

145.82 + F = 255.06

F = 255.06 - 145.82 = 109.24 N

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3 years ago
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