45 degrees below the ground

Classical Mechanics Level 3

A projectile is thrown horizontally from a height h h and lands at a horizontal distance d d from where it's thrown. If, at the landing point, the velocity vector of the projectile makes an angle of 4 5 ∘ 45^\circ below the ground, what is the value of d h \frac{d}{h} ?

1 2 4 1/2

Erasmo Hinojosa by Erasmo Hinojosa , 30, Mexico

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1 solution

Erasmo Hinojosa
May 6, 2016

This problem is related to the value of the velocity components in a given point. Since the vertical displacement of the projectile can be modeled by:

y = g t 2 2 → t = 2 y g y=\frac{gt^2}{2} \rightarrow t=\sqrt{\frac{2y}{g}}

So, the magnitude of the velocity's components at the landing point is given by:

v x = d g 2 h ; v y = 2 g h v_x=d \sqrt{\frac{g}{2h}} ; v_y=\sqrt{2gh}

At the landing point, the velocity vector makes an angle of 45º below the ground, meaning that the magnitude of the velocity components should be the same:

d g 2 h = 2 g h → d h = 2 d \sqrt{\frac{g}{2h}}=\sqrt{2gh} \rightarrow \frac{d}{h}=2

Thus, the anwser is 2.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

My solution:

Since the angle is 4 5 ∘ 45^{\circ} we know that the vertical and horizontal components of the velocity are equal in magnitude.
∣ V ⃗ x ∣ = ∣ V ⃗ y ∣ \left| \vec{V}_{x} \right| = \left| \vec{V}_{y} \right|
V ⃗ y = ∫ ( a ⃗ ) d t = ∫ ( − g ) d t \vec{V}_{y} = \displaystyle \int (\vec{a}) \text{d}t = \displaystyle \int (-g) \text{d}t
V ⃗ y = − g t \vec{V}_{y} = -gt
h = ∫ ( − g t ) d t h = \displaystyle \int (-gt) \text{d}t
h = − g t 2 2 h = \dfrac{-gt^{2}}{2}
V ⃗ x = d t = − g t \vec{V}_{x} = \dfrac{d}{t} = -gt
d = − g t 2 d = -gt^{2}
d h = − g t 2 − g t 2 2 = 2 \dfrac{d}{h} = \dfrac{-gt^{2}}{\dfrac{-gt^{2}}{2}} = 2

A Former Brilliant Member - 5 years ago

My alternative solution:

Let us use a coordinate grid system with y axis being height and x axis being horizontal displacement.

We know that the equation of x and y are as follows: y = − g 2 t 2 + h x = v t y=-\frac{g}{2}t^2+h\\x=vt

Combining the two, we have y = − g 2 v 2 x 2 + h y=-\frac{g}{2v^2}x^2+h

We can find d d by setting y to 0. 0 = − g 2 v 2 d 2 + h d = v 2 h g 0=-\frac{g}{2v^2}d^2+h \\d=v\sqrt{\frac{2h}{g}}

Then, using the other condition in the question, we have d y d x = − g v 2 h g = − 1 g 2 v 2 × 2 h g = 1 2 h g = v 2 h = v 2 2 g \frac{dy}{dx}=-\frac{g}{v}\sqrt{\frac{2h}{g}}=-1\\\frac{g^2}{v^2}\times\frac{2h}{g}=1\\2hg=v^2\\h=\frac{v^2}{2g}

Now, we set it back to the equation of d. d = v v 2 g 2 = v 2 g d=v\sqrt{\frac{v^2}{g^2}}=\frac{v^2}{g}

Combining the two, we can clearly see that d h = 1 1 2 = 2 \frac{d}{h}=\frac{1}{\frac{1}{2}}=2

A Former Brilliant Member - 5 years ago

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Great solution! More formal than mine :P

Erasmo Hinojosa - 5 years ago

Nice solution! However, there is a slight typo in the last line; it should be d h \frac dh .

Shaun Leong - 5 years ago

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Noted with thanks.

A Former Brilliant Member - 5 years ago

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