# a body is projected with velocity u such that its horizontal range and maximum vertical heights are

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## A body is projected with velocity u such that its horizontal range and maximum vertical heights are same. The maximum heights is:

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## A body is projected with velocity u such that its horizontal range and maximum vertical heights are same. The maximum heights is:

**A**

2g u 2

**B**

4g 3u 2

**C**

17g 16u 2

**D**

17g 8u 2 Medium Open in App Solution Verified by Toppr

Correct option is D)

Lets consider u= initial speed of the projectileθ= angle of projection

According to question,

H=R 2g u 2 sin 2 θ = g u 2 sin2θ

sinθ.sinθ=2×2sinθ.cosθ

sinθ=4cosθ tanθ=4

As we know that, 1+tan

2 θ=sec 2 θ cos 2 θ 1 =1+16=17 cos 2 θ= 17 1 sin 2 θ+cos 2 θ=1 sin 2 θ=1−cos 2 θ sin 2 θ=1− 17 1 = 17 16 Maximum height, H= 2g u 2 sin 2 θ H= 2g u 2 17 16 H= 17g 8u 2

The correct option is D.

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## A body is projected with velocity u such that in horizontal range and maximum vertical heights are samek.The maximum height is

:.Tan theta=4 H(max)=(u^(2)sin^(2)theta)/(2g)A body is projected with velocity u such that in horizontal range and maximum vertical heights are samek.The maximum height is

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A body is projected with veloc...

A body is projected with velocity

u u

such that in horizontal range and maximum vertical heights are samek.The maximum height is

Updated On: 27-06-2022

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Text Solution Open Answer in App A u 2 2g u22g B 3 u 2 4g 3u24g C 16 u 2 17g 16u217g D 8 u 2 17g 8u217g Answer

The correct Answer is D

Solution ∴Tanθ=4 ∴Tanθ=4 H max = u 2 sin 2 θ 2g Hmax=u2sin2θ2g Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6

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## A body is projected with velocity u such that its horizontal class 11 physics CBSE

A body is projected with velocity u such that its horizontal range and maximum vertical heights are the same The maximum heights is A dfracu22g B dfrac3u24g C dfrac16u217g D dfrac8u217g

A body is projected with velocityu u

such that its horizontal range and maximum vertical heights are the same. The maximum heights is:

A. u 2 2g u22g B. 3 u 2 4g 3u24g C. 16 u 2 17g 16u217g D. 8 u 2 17g 8u217g Answer Verified 105.3k+ views

**Hint:**Initially, we write down the given information. Then we equate the formulas for horizontal range and the maximum height to each other. This allows us to find the value of sine of the angle of projection. We apply this value of sine of the angle of projection to the formula of maximum height and find the maximum height of the motion.

**Formulas used:**

The maximum height of projectile motion is given by the formula,

H= u 2 sin 2 θ 2g H=u2sin2θ2g

The horizontal range of the projectile motion is given by the formula,

R= u 2 sin2θ g R=u2sin2θg Where θ θ

is the angle of projection,

u u

is the initial velocity of the projectile and

g g

is the acceleration due to gravity.

**Complete step by step answer:**

It is given that u u

is the initial velocity of the projectile and the maximum vertical height the projectile can reach is equal to the horizontal range of the projectile.This gives us,

u 2 sin2θ g = u 2 sin 2 θ 2g u2sin2θg=u2sin2θ2g

Cancelling the like terms from both sides, we get

sin2θ= sin 2 θ 2 sin2θ=sin2θ2

We simplify this equation and arrive at,

2sinθcosθ= sinθ×sinθ 2

2sinθcosθ=sinθ×sinθ2

We cancel the sine value and the remainder will be,

2cosθ= sinθ 2 2cosθ=sinθ2

Taking the trigonometric functions to one side and the numerals to the other,

4= sinθ cosθ 4=sinθcosθ

But we know that this is the tangent. That is,

tanθ=4 tanθ=4 .

Using the trigonometric identity,

1+ tan 2 θ= sec 2 θ 1+tan2θ=sec2θ .

We can find the value of secant squared and taking the reciprocal, we find the value of cosine squared

sec 2 θ=17 sec2θ=17

and we take the reciprocal to get,

cos 2 θ= 1 17 cos2θ=117

We apply another trigonometric identity,

sin 2 θ+ cos 2 θ=1 sin2θ+cos2θ=1

We get the value of sine squared as,

sin 2 θ=1− cos 2 θ ⇒ sin 2 θ= 17−1 17 ⇒ sin 2 θ= 16 17

sin2θ=1−cos2θ⇒sin2θ=17−117⇒sin2θ=1617

Substituting this value in the equation to find the maximum height, we get

H= u 2 sin 2 θ 2g ⇒H= u 2 ×16 2×17×g ∴H= 8 u 2 17g

H=u2sin2θ2g⇒H=u2×162×17×g∴H=8u217g

**Hence, the right answer is option D.**

**Note:**In projectile motion, the vertical component of velocity is zero or absent. This is because, as the projectile moves upward the movement taking place will be against gravity and this makes the projectile to decelerate.

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