Question

Mt. Asama, Japan, is an active volcano complex. In 2009, an eruption threw solid volcanic rocks that landed far from the crater. Suppose that one such rock was launched at an angle of θ = 58.7 degrees above horizontal, and landed a horizontal distance d = 1200 m from the crater, and a vertical distance h = 780 m below the crater.
Write and expression for v0, the initial speed of the rock in terms of θ, d, and h.

Answer 1

Answer:u=97.41m/s

Explanation:

Given

inclination $\theta =58.7^{\circ}C$

Horizontal distance travel by Particle $d=1200 m$

Vertical height $h=780 m$

Let u be the initial velocity

calculating vertical distance

$y=u\sin \theta +\frac{at^2}{2}$

$y=u\sin \theta t-\frac{gt^2}{2}$-------1

Calculating horizontal distance

$x=u\cos \theta \times t+0$

$t=\frac{x}{u\cos \theta }$

put value of t in equation 1

$y=u\sin \theta \times \frac{x}{u\cos \theta }-\frac{g}{2}\times (\frac{x}{u\cos \theta })^2$

$y=x\tan \theta -\frac{gx^2}{2u^2\cos ^2\theta }$

$\frac{gx^2}{2u^2\cos ^2\theta }=x\tan \theta -y$

$u^2=\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}$

$u=\sqrt{\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}}$

at $y=-780\ m\ x=1200 m$

$u^2=\frac{18.154}{2753.65}\times 1200^2$

$u=97.41 m/s$