Physics 20-1 Final Review: Projectiles

[physicznoob]

Homework Statement

This question is something we got in our review booklet and I HATE projectile motion.

An object is thrown horizontally with a velocity of 18 m/s from the top of a cliff. If the object hits the ground 100 m from the base of of the cliff, how high is the cliff?

Homework Equations

d = v1t + 1/2at^2, v = d/t

The Attempt at a Solution

I try to break it up into components(vertical and horizontal) but nothing seems to work.

I used the equation d = 1/2v1 + 1/2at^2 to solve for t, because obviously the initial velocity is 0 m/s, but i end up with a far-fetched answer.

 

[amd123]

for some reason i have the same question, if you help me in my two threads, ill help you with yours. I'm beast at projectile motion

 

[physicznoob]

so we're bargaining now? lol.

 

[bp_psy]

1.You don't have any acceleration in the x direction so you can use that to find t.
The fall ends when the object has traveld 100 m in the x direction at a constant velocity
2. The equation is not correct d=v₀t+(1/2)a*t²

 

[physicznoob]

i still don't get it

 

[amd123]

okay its thrown horizontally so no Vi in the Y direction
g is -9.8
you have distance in the x direction for 100m
you need the velocity in the x direction and now you solve for time vx = 18ms
T = dx/vx so 100/18 = T in seconds

then now solve for distance in the y direction
D = viy*t +1/2(-9.8)(t^2) viy = 0 so solve for T and then solve for D

i think this should be right but I am not sure

now you got to help me more I am still confused :(

 

[wizard85]

You can use a frame of reference $$(X,Y)$$ with origin $$(0,0)$$ in the cliff and the point $$P(0,s_{oy})$$ is where the projectile will be thrown.
Equation of motion respect to X and Y are:$$\begin{align*} s_x=v_ox*t + s_{ox}\\ s_y=\frac{1}{2} * g * t^2 + v_{oy}*t+s_{oy} \end{align*}$$but in according to my frame of reference $$s_{ox}=0$$. Nevertheless $$v_{ox}=0$$ and $$v_{oy}=0$$. You can make the equation of motion explicit in the following way (t disappeared):

$$s_y=\frac{1}{2} * g * \frac{s_x^2}{v^2} + s_{oy}$$

but where the projectile falls in ground

$$s_y=0$$

then:

$$- \frac{1}{2} * g * \frac{s_x^2}{v^2} = s_{oy}$$

if you replace $$s_x$$ and $$v$$ given by the problem you will have $$s_{oy}$$ which is the height of the cliff from which the stuff is thrown (if I have done no error it should be approximately 151.39 m).