Question

jee mains related question​

Answer 1

Given

• Initial velocity = u
• Angle of elevation = θ wih the horizontal

To Find

• Relation between the maximum height (H) & range (R)

Solution

☯ R = u²sin2θ/g

☯ H = u²sin²θ/2g

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✭ According to the Question :

➞ R = u²sin2θ/g

➞ R = u²sin2θ × 1/g

➞ R/sin2θ = u²/g -eq(1)

Similarly,

➞ H = u²sin²θ/2g

➞ H/2sin²θ = u²/g -eq(2)

Equating eq(1) & eq(2)

➞ R/sin2θ = H/2sin²θ

➞ R/sinθ.cosθ = H/2 × sinθ × sinθ

➞ R/2cosθ = 2H/sinθ

➞ R × sinθ = 2H × 2cosθ

➞ Rsinθ = 4Hcosθ

➞ H = Rsinθ/4cosθ

➞ H = Rtanθ/4

∴ The answer is Option C

Answer 2

Answer:

h = (R tanθ)/4

Explanation:

From the three equations of motion we get

$Range =\frac{ 2u^{2} sin\theta cos\theta}{g} \\Height (max) = \frac{u^{2}sin^{2}\theta }{2g}$

Now to find h in terms of R we need to divide the two formulas

$\frac{R}{h} = \frac{2u^{2}sin\theta cos\theta }{g} \times \frac{2g}{u^{2}sin^{2} \theta}$

on simplifying this we get

$\frac{R}{h} = 4cot\theta$

∴ $h= \frac{Rtan\theta}{4}$

took me way too long to type, hope it helps :)