A stone is thrown off a cliff...

[TheDragon]

1. The problem statement, all variables and given/known data
A stone is thrown off of a bridge that is 40m above the water. The stone lands 48m from the base of the bridge. The stone was thrown with an initial velocity of 15m at 30 degrees above the horizontal.

What is the maximum height the stone reaches above the water? (ans:42.87m)

2. Relevant equations
 

•  
  • Vfx=Vix=VicosΘ
  • Δx=vicosΘ(t)
  • Vfy=(VicosΘ)+ay(t)
  • Δy=(VisinΘ)(t)+1/2(ay)(t)
  • (Vfy)^2=(VisinΘ)^2+2(ay)(Δy)
  • Δy=1/2(Vfy+(VisinΘ))(t)
  • a=0m/s in x-direction

3. The attempt at a solution
ax:13
ay:7.5
Δy:40m
Δx:48m

I found the x and y-components. Ax:15cos30=13 and Ay:15sin30=7.5.

Then I tried to find the time using the second equation and got 3.69s.

I've been trying to solve for vfy.... and I've done everything. I'm not getting that answer. I don't know why I can't understand how to do this. it's frustrating... I even tried to draw it on an axis.

 

 

 

[studiot]

Does this help?

 

How did you get on with your other projectile problem?

Edited September 7, 2018 by studiot

[TheDragon]

i found the components. 

ax=13m/s and ay=7.5m/s

but i didnt care for the x. The viy is 7.5 and the vfy is 0 then i subbed into the equation.

vfy^2-7.5^2=2(-9.8)changeinY

-56.25=-19.6changeinY

I divided on both sides and got 42.87m 

 

The next part of the equation is asking for the final vleocity before it hit the water. I'm not sure how to do that.

[studiot]

14 hours ago, TheDragon said:

it's frustrating...

 

It's also frustrating when you don't seem interested in discussion.