Question

Worked out problems: Solve completely algebraically before inserting any values. Make sure your algebraic answer is in terms of given variables 5. A rock is thrown over a cliff from the origin with initial velocity +z points in the upward direction. where is it a time later? Ve,+o where A. (4 points) List two assumptions that you need to make to solve this projectile motion problem. B. (8 points) For the initial velocity, = 3 Isi-24, +2.02; where is the rock 2.0 s later? C. (8 points) If the ground is 65 m below the cliff, when will the rock hit the ground? Use the same initial velocity as in part B.
Need some help.

Answer 1

5) XY plane is horizontal and Z-axis is pointing upward.

Initial velocity of rock $\vec{v}_0=v_{ox}\hat{x}+v_{oy}\hat{y}+v_{oz}\hat{z}$

After time $\Delta t$ ,

The X-coordinate of rock is $x(t)=v_{ox}\Delta t+\frac{1}{2}a_x(\Delta t)^2$ ​

The Y-coordinate of rock is $y(t)=v_{oy}\Delta t+\frac{1}{2}a_y(\Delta t)^2$ ​

The Z-coordinate of rock is $z(t)=v_{oz}\Delta t+\frac{1}{2}a_z(\Delta t)^2$

At time $\Delta t$ , the position of particle is $\vec{r}=\vec{v}_0 \Delta t +\frac{1}{2}\vec{a}(\Delta t)^2$

A) The two assumptions are,

Acceleration is constant and acting vertically downwards , There is no air resistance in the rock's motion.

Since X and Y-axes are horizontal (and perpendicular to each other) , $a_x=0$ , $a_y=0$ and $a_z=-g$

So the velocity along X-axis and velocity along Y-axis do not change with time.

Velocity along Z-axis changes with time since the acceleration along Z-axis is changing.

Position coordinates of the rock after time $\Delta t$ ,

$x=v_{ox}\Delta t$ , $y=v_{oy}\Delta t$ ,   $z=v_{oz}\Delta t-\frac{1}{2}g(\Delta t)^2$

B)

$\vec{v}_0=(3.1\hat{x}-2.4\hat{y}+2.0\hat{z})\,m/s$ $\Delta t=2.0\,s$

$x=v_{ox}\Delta t=3.1*2.0=6.2\,m$

$y=v_{oy}\Delta t=-2.4*2.0=-4.8\,m$

$z=v_{oz}\Delta t-\frac{1}{2}g(\Delta t)^2=2.0*2.0-\frac{1}{2}*9.81*(2.0)^2=-15.62\,m$

$(x,y,z)=(6.2,-4.8,-15.62)\,m$

$\vec{r}=(6.2\hat{x}-4.8\hat{y}-15.62\hat{z})\,m$

C) Given that $Z=-65\,m$ ( downward vectors are taken negative)

$z=v_{oz}\Delta t-\frac{1}{2}g(\Delta t)^2$

$-65=2.0\,t-\frac{1}{2}*9.81 \,t^2$

$4.904\,t^2-2\,t-65=0$

solving above quadratic equation $t=3.9\,s$