Projectile Motion Off a Cliff: Solving for Minimum Speed and Distance Traveled

[swanny1286]

Homework Statement

A 88.0 kg boulder is rolling horizontally at the top of a vertical cliff that is 20 m above the surface of a lake, as shown in the figure . The top of the vertical face of a dam is located 100 m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25m below the top of the dam. (height=h=45m, distance=deltaX=100m)

A)What must be the minimum speed of the rock just as it leaves the cliff so it will travel to the plain without striking the dam?
B)How far from the foot of the dam does the rock hit the plain?

Homework Equations

(1) Vf^2-Vo^2=2*a,y*h
(2) Vf-Vo=a*t
(3) Yf=Yo + Vo,y(t) +.5*a*t^2
(4) Xf,x=Xf + Vo,x(t) + .5*a,x*t

The Attempt at a Solution

I had figured the origin would be at the top of the dam. using the eqn 3, with Yo= 0 and Vo,y=0, i could use that to solve for time: t=sqrt((2*h)/g)= 3.0secs. with time i could find the Vo,x because it states how far over the dam it must fly.

My original answer of part A was incorrect of 33.0 m/s and without it i have no clue how to set up B

 

[rl.bhat]

Hi swanny1286,
Welcome to PF.
What is time taken by the boulder to fall vertically through 20 m.
If it is t = sqrt(2*h/g) = 3.0 sec, check this calculation.

 

[noblegas]

Homework Statement

A 88.0 kg boulder is rolling horizontally at the top of a vertical cliff that is 20 m above the surface of a lake, as shown in the figure . The top of the vertical face of a dam is located 100 m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25m below the top of the dam. (height=h=45m, distance=deltaX=100m)

A)What must be the minimum speed of the rock just as it leaves the cliff so it will travel to the plain without striking the dam?
B)How far from the foot of the dam does the rock hit the plain?

Homework Equations

(1) Vf^2-Vo^2=2*a,y*h
(2) Vf-Vo=a*t
(3) Yf=Yo + Vo,y(t) +.5*a*t^2
(4) Xf,x=Xf + Vo,x(t) + .5*a,x*t

The Attempt at a Solution

I had figured the origin would be at the top of the dam. using the eqn 3, with Yo= 0 and Vo,y=0, i could use that to solve for time: t=sqrt((2*h)/g)= 3.0secs. with time i could find the Vo,x because it states how far over the dam it must fly.

My original answer of part A was incorrect of 33.0 m/s and without it i have no clue how to set up B

Your equation (1) is wrong ; Should be v^2-v_0^2=2*a*y

 

[swanny1286]

OH ok, so the boulder just has to clear the dam, so only falling 20 m instead of 45 m. it only takes 2s. so finding that, and knowing the final distance, i could solve for the initial velocity. now i am just wondering how to approach part B because I am not sure what to start at.

i do have a feeling that using the inital velocity, i can find how fast it's going to be going (in components) right past the dam and find how far it flies from there (ofc adding the length of the lake) or am i off ?

edit: with going back and re-reading the question, i had found that you only need to know how far it's going to fly and that the length of the dam will be included, seeing how i just found that for part A, so its Xf-100.

I had gone through and calculated, just like in A, for the time that it flies and then using that with the initial X velocity to find the distance (~147m)-100m for the dam and i had gotten it wrong. is my thought process off?

 

[swanny1286]

Your equation (1) is wrong ; Should be v^2-v_0^2=2*a*y

my habit of writing its is V(final)^2-V(initial)^2, so i just use the Vf instead of just V, unless i am wrong and it's not the finally velocity and just the magnitude?

 

[rl.bhat]

In 2 seconds the boulder must cover 100m. What is the initial velocity.
Now how much it covers in 3 seconds?

 

[swanny1286]

In 2 seconds the boulder must cover 100m. What is the initial velocity.
Now how much it covers in 3 seconds?

Thats what i had assumed for the second part, that if it took 2s to cover that distance @ 49m/s, and then it went for 1 more second it would go 147 meters from the base of the cliff, 47 m from the base of the dam, but it is incorrect...

 

[rl.bhat]

Thats what i had assumed for the second part, that if it took 2s to cover that distance @ 49m/s, and then it went for 1 more second it would go 147 meters from the base of the cliff, 47 m from the base of the dam, but it is incorrect...

I am getting 50 m.
What is the expected answer?

 

[swanny1286]

I am getting 50 m.
What is the expected answer?

so uh apparently it was 50m, not 47m. it was a rounding error on my part... thanks a bunch for your help, i now throughly understand it!