Answer to Question #158285 in Classical Mechanics for Chadi A’ishah

Question

A rock ejected from a volcano with a speed of 25m/s and at an angle of 35 degrees above the horizontal. If the rock hits the side of the volcano at an altitude 20m lower than the starting point, how long does it take the rock to reach the starting point?

Answer 1

Given,

Speed of ejection of rock from the volcano $(v)= 25m/s$

Angle $(\theta) = 35^\circ$

Height below to the volcano at which rock get $(h)=20m$

Now, taking the vertical and horizontal component of velocity,

Velocity along x axis $v_x=v\cos\theta=25\sin 35^\circ$

Velocity along y axis $v_y = v\sin \theta= 25\sin 35^\circ$

The horizontal component does not change with time but the vertical component of the velocity will vary because there is no any acceleration along the x axis but in the vertical direction or along to the y axis, there is force of gravity is working,

gravitational acceleration $(g)=9.8 m/s^2$

$-h = v_y t-\frac{gt^2}{2}$

Now, substituting the values,

$\Rightarrow -20 = 25\sin(35^\circ)t-\frac{9.8\times t^2}{2}$

$\Rightarrow -20 = 14.34t-4.9t^2$

$\Rightarrow 4.9t^2-14.34t-20=0$

$\Rightarrow t=\frac{14.34\pm \sqrt{14.34^2+4\times 4.9\times20}}{9.8}$

$\Rightarrow t=\frac{14.34\pm24.44}{9.8}$

Now, taking the '+'

$t=\frac{14.34+24.44}{9.8}s$

$\Rightarrow t = \frac{38.78}{9.8}$ s

$\Rightarrow t = 3.95s$

taking $'-'$

$t=\frac{14.34-24.44}{9.9}=\frac{-10.4}{9.8}s$

time can not be negative,

hence we will ignore this value.

Hence time taken by the rock to reach the 20 m below height be 3.95 second