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(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

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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.

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.

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