A ball is projected from the ground with a speed 15 ms^-1 at an angle θ with horizontal so that its range and maximum height are equal. Then 'tanθ' will be equal to

Option 1 - <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>

Option 3 - 2

Option 4 - 4

2 Views|Posted 6 months ago

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Answered by

Payal Gupta | Contributor-Level 10

6 months ago

Correct Option - 4

Detailed Solution:

$H = \frac{u^{2} s i n^{2} θ}{2 g} a n d R = \frac{u^{2} s i n 2 θ}{g}$

According to question

$R = H \Rightarrow \frac{u^{2} s i n^{2} θ}{2 g} = \frac{2 u^{2} s n θ c o s θ}{g} \Rightarrow t a n θ = 4$

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According to question, we can write

$R = \frac{u^{2} s i n 2 θ}{g} \Rightarrow u^{2} = g R_{m a x}, w h e r e θ = 45 °$

$\Rightarrow H_{m a x} = \frac{u^{2}}{2 g} = \frac{g R_{m a x}}{2 g} = \frac{R_{m a x}}{2} = \frac{100}{2} = 50 m$

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A ball is projected from the ground with a speed 15 ms-1 at an angle θ with horizontal so that its range and maximum height are equal. Then 'tanθ' will be equal to

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A ball is projected from the ground with a speed 15 ms^-1 at an angle θ with horizontal so that its range and maximum height are equal. Then 'tanθ' will be equal to