Question

Two stone are projected with same speed but making different angles with the horizontal. Their ranges are equal. If the angle of projection of one is 60 degree and it's maximumheight is y1 then the maximum height of the other will be:-
a. 3y​1 b. 2y1 c. y1/2 d. y1/3

I want the answer with full solution.

Answer 1

Range is same for complementary angles as,
range= u²sin2ø/g,
and, sin2(90-ø)= sin(180-2ø)= sin2ø.
So, the angle for the second stone will be 30°.
y1= u²sin²(60°)/2g and y2= u² sin²(30°)/2g.
From both equations, y2= y1/3.

Answer 2

Answer:

The maximum height of the other stone is y1/3.

Projectile Motion:

• An object thrown in the air with some initial velocity in any direction, making some angle with the horizontal, moving freely under the action of gravity is called a projectile.
• The motion of the object is called as projectile motion.
• The velocity with which the body is projected is called the velocity of projection.
• The horizontal distance between the point of projection and the point on the same horizontal plane, at which the projectile returns after moving along its trajectory, is called the range of the projectile.
• The maximum vertical distance traveled by the projectile from the ground level during its motion is called the height of the projectile.

Explanation:

Given:

$u_{1}=u_{2}$, θ1 = 60°, $H_{max1} = y1$

To find:

Maximum height of the other stone ($H_{max2}$)

Formula:

$R=(\frac{u^{2} }{g})$ × (sin2θ)

$H=(\frac{u^{2} }{2g} )$ × ($sin^{2}$θ)

Step 1:

The range of both the projectiles is equal. Therefore, $R_{1}=R_{2}$

$(\frac{u_{1} ^{2} }{g} )$  × (sin2(θ1)) = $(\frac{u_{2} ^{2} }{g} )$ × (sin2(θ2))

Since $u_{1}=u_{2}$ we get

sin2θ1 = sin2θ2

Substituting the value of θ1 = 60°

sin 2(60°) = sin 2(θ2)

sin 120° = sin 2(θ2)

0.8660 = sin 2(θ2)

2(θ2) = $sin^{-1}(0.8660)$

2(θ2) = 60°

θ2 = 30°

Step 2:

For one stone, the maximum height is given to be y1 and the angle is 60°.

Substituting these values in the formula for height, we get

$H=(\frac{u^{2} }{2g} )$ × ($sin^{2}$θ)

$y1=(\frac{u^{2} }{2g} )$ × ($sin^{2}$60)

$y1=(\frac{u^{2} }{2g} )$ × $(0.8660)^{2}$

$y1=(\frac{u^{2} }{2g} )0.75$ ...(i)

Step 3:

Let y2 be the maximum height of the other stone. The angle of projection for this stone is 30°.

Substituting these values in the formula for height, we get

$H=(\frac{u^{2} }{2g} )$ × ($sin^{2}$θ)

$y2=(\frac{u^{2} }{2g} )$ × ($sin^{2}$30)

$y2=(\frac{u^{2} }{2g} )$ × $(0.5)^{2}$

$y2=(\frac{u^{2} }{2g} )0.25$ ...(ii)

Step 4:

Substituting equation (ii) in equation (i), we get

$y1 = 3 [(\frac{u^{2} }{2g})0.25]$

$y1=3(y2)$

$y2=\frac{y1}{3}$

Therefore, the maximum height of the other stone in terms of y1 will be y1/3.

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