مساعدة في حل أسئلة عن حركة المقذوفات

[إحساسي كبير]

السلام عليكم ورحمة الله و بركاته
أتمنى بس مساعدتي بحلول هذه الأسئلة الثلاثة

http://im42.gulfup.com/Jqoq0.jpg

و

http://im42.gulfup.com/kGVzl.jpg

وجزاكم الله خيرا وجعله في ميزان حسناتكم

[ام عبد الرحمن محمد]

(a) Range of a projectile is given by
R = (u^2 sin 2θ) / g
The maximum height is given by
H = (u^2 sin^2 θ) / 2g

Here R = H
Simplify to get
sin 2θ = sin^2 θ / 2
Put sin 2θ = 2 sin θ cos θ
So
4 cos θ = sin θ
or tan θ = 4

θ = tan^-1 (4) = 76 °

(b) The answer does not depend on acceleration due to gravity (g); it cancels out in the beginning. So the answer would be same even on another planet.

(c) R = (u^2 sin 2θ) / g
Greater the value of sin 2θ, greater the range.
sin 2θ lies between -1 and 1.
So maximum value of sin 2θ = 1.
When sin 2θ = 1,
2θ = 90
or θ = 45

Put sin 2θ = 1 in the expression for Range :
R = u^2 / g

[خلف الجميلي]

هل من الممكن تعطينا اسم الكتاب ؟

[ام عبد الرحمن محمد]

السؤال 13
The time to travel 36 m is given by
t = 36 / 20 cos 53 = 2.991 s.

In this time it travels a vertical height given by
h = 20 sin 53*2.991 – ½ * 9.8 *2.991^2
h =3.94 m

Since 3.94 m is more than 3.05 m
the ball is above the cross bar by a height of 0.89.

The velocity at the end of 2.991 s is
v = u –gt = 20 sin 53 – 9.8*2.991 = -13.34.
The negative sign shows that the ball is falling down
while approaching the cross bar.

[ام عبد الرحمن محمد]

سؤال 14
How far horizontally from the base of the building does the ball strike the ground? >>

X = Vx(T)

where

X = horizontal distance from base of building where ball will land
Vx = horizontal component of velocity
T = time ball strikes the ground after being tossed

Substituting appropriate values,

X = 8(cos 20)(3)

X = 22.55 m


<< Fund the height from which the ball was thrown. >>

Y = Vy(T) + (1/2)(g)T^2

Y = height from which ball was thrown
Vy = vertical component of the velocity
T = time when ball strikes the ground
g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting appropriate values,

Y = 8(sin 20)(3) + (1/2)(9.8)(3^2)

Y = 52.31 meters


<< How long does it take the ball to reach a point 10.0 m below the level of launching? >>

Formula is

S = Vy(T) + (1/2)(g)T^2

where

S = 10 m
Vy = 8(sin 20)
g = 9.8 m/sec^2
T = time for ball to reach a point 10 m below the launching level

Substituting values,

10 = 8(sin 20)T + (1/2)(9.8)T^2

Rearranging the above,

1/2)(9.8)T^2 + 8(sin 20)T - 10 = 0

Solving for "T" using the quadratic formula,

T = 0.22 sec.

[خلف الجميلي]

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