Question

Post - Laboratorio Movimiento en dos Dimensiones - Proyectiles Nombre: Sección: 1. ¿A qué ángulo usted obtuvo el mayor alcance en la segunda parte del laboratorio? ¿A qué án- gulo acurre el mayor alcance según la ecuación (5.5)? ¿Sus resultados son consistentes con esta predicción teórica? En la segunda parte del laboratorio se obtuvo que el angulo (45°) obtuvo el mayor alcance. Según la ecuacon (5.5), el ángulo de mayor alcance ocurre a los 2. En el caso de lanzamiento horizontal, ¿Qué tendría mayor resultado para aumentar el alcance, doblar la velocidad inicial del proyectil o doblar la altura a a que cae el proyectil? Explique.

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Answer 1

1). The equation for range of the object underground projectile motion is given as

“s |- * sin(20)

The angle $\theta$ varies from 0 to 90 °. So the
Sin (20)
varies from 0 to 1. That means for the range to be maximum,
Sin (20)
must be 1. And for that to happen,$2\theta$ must be 90.

Therefore $\theta$ must be 45°

2). Let us consider the horizontal launch now.

The horizontal component of projectile velocity remains invariant throughout it's motion but the vertical component changes.

Therefore the horizontal range therefore

X = vot

$t=\frac{X}{v_0}$

And the vertical distance travelled during the same time can be calculated as follows.

Assume that the height of projectile is y_{​​​​​​0}​​​

The projectile flies for time t.

Therefore the distance covered by it vertically is

. Where v_{​​​​​​0y}​ is initial velocity in y direction which is zero.

h = voy + 5gt

જે?

The negative sign indicates that the projectile is moving downwards.

So the height above the ground vertically is given as

. When this becomes zero, that's when the projectile heats the ground. But we have

y = ho - gt

$t=\frac{X}{v_0}$

$y=h_0-\frac{1}{2}(\frac{X}{v_0})^2$​​​​​​

Now the quantity y-h_{​​​​0} must give the total vertical distance travelled by the projectile since h_{​​​​​​0}​​​​​ is the maximum height and y is the distance yet to be travelled at any instance.

$Y=y-h_0=\frac{1}{2}(\frac{X}{v_0})^2$............(a)

So if the height of launch is doubled we get

$2Y=\frac{1}{2}(\frac{X_1}{v_0})^2$..............(b)

The ratio of eqn (b) to (a) gives

U2 22

$\therefore X_1=\sqrt{2}X$

Now if we double the initial velocity v_{​​​​​​0}​​​ we get

$Y=\frac{1}{2} \times (\frac{X}{2v_0})^2$

There the range increases 4 times. Thus it is more effective to double the velocity