Shooting a deer can be a tough job!

Classical Mechanics Level 5

A hunter in a valley is trying to shoot a deer on a hill. The distance of the deer along his line of sight is 10 181 m 10\sqrt{181} \text{ m} and the height of the hill is 90 m 90\text{ m} . His gun has a muzzle velocity of 100 ms − 1 100\text{ ms}^{-1} . Find maximum how many meters above the deer he should aim his rifle in order to hit it.

Details and Assumptions

Take acceleration due to gravity as 10 ms − 2 10\text{ ms}^{-2} .


The answer is 1810.

Akshat Sharda by Akshat Sharda , 21, India

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2 solutions

Akshat Sharda
Jun 14, 2016

By using equation of trajectory,

y = x tan ⁡ θ − g x 2 2 u 2 cos ⁡ 2 θ y=x\tan \theta -\frac{gx^2}{2u^2\cos^{2}\theta}

Here, θ \theta is the angle at which the hunter should aim above the deer in order to hit it.

90 = 100 tan ⁡ θ − 10 ⋅ 10 0 2 ( 1 + tan ⁡ 2 θ ) 2 ⋅ 10 0 2 tan ⁡ 2 θ − 20 tan ⁡ θ + 19 = 0 ( tan ⁡ θ − 19 ) ( tan ⁡ θ − 1 ) = 0 tan ⁡ θ = 1 , 19 90=100\tan \theta -\frac{10\cdot 100^2(1+\tan^2 \theta) }{2\cdot 100^2} \\ \tan^2 \theta -20\tan \theta +19=0 \\ (\tan \theta-19)(\tan \theta -1)=0 \\ \tan \theta =1,19

Maximum value, ⇒ 1900 − 90 = 1810 \Rightarrow 1900-90=\boxed{1810}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

But isnt there something different we cant have t = 19 t=19 because it will miss the deer by a jiffy then. So, the maximum distance is a bit less that 1 2 × 10 × 19 × 19 = 1805 \dfrac{1}{2} × 10 × 19×19 = 1805 which is 1804.99...... 1804.99...... where am I going qrong?

Ashish Menon - 5 years ago

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Why will it miss the deer? Can you please elaborate?

Akshat Sharda - 5 years ago

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Wait, that concept applies for minimum time!!! Ah, I made a silly mistake :( nice question :)

Ashish Menon - 5 years ago

A question of Resonance sheet.

A Former Brilliant Member - 4 years, 10 months ago

Please can you explain the last step? How are you getting the maximum value because there is no way I am getting this right.

Why can we differentiate the equation of trajectory and find the value of x for maximum y and apply that in the equation to find maximum vertical height and then subtract 90 from it?

Somrat Dutta - 3 years, 11 months ago

Let θ \theta be the angle that the hunter should aim at to hit the deer. We know that the maximum angle occurs when the bullet hits the deer during its descent. Also, let t t be the time that the bullet stays in the air. Hence, we have 90 = 100 sin ⁡ θ t − 5 t 2 t = 10 sin ⁡ θ + 100 sin ⁡ 2 θ − 18 100 cos ⁡ θ ⋅ t = 100 cos ⁡ θ ( 10 sin ⁡ θ + 100 sin ⁡ 2 θ − 18 ) = 1 θ ≈ 86.987 90=100\sin\theta t-5t^2\\t=10\sin\theta+\sqrt{100\sin^2\theta-18}\\\\100\cos\theta \cdot t=100\\\cos\theta(10\sin\theta+\sqrt{100\sin^2\theta-18})=1\\\theta\approx86.987 .

The distance that the hunter should aim above the deer is then 100 tan ⁡ θ − 90 ≈ 1810 100\tan\theta-90\approx1810

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Why angle can't be 45 degree

Arun Krishna AMS - The Joker - 4 years, 11 months ago

90 = 100sin©t - 5t^2 100 = 100 cos© t Then t = 1/cos© substituting this in first equation 90 = 100 sin©/cos©- 5(1/cos©)^2 If we substitute sin© = cos© =( 1/(2)^(1/2)) We get both LHS and RHS the same Thus 45degree can also be the angle

Arun Krishna AMS - The Joker - 4 years, 11 months ago

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