A stone is thrown from a cliff

Question

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.

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.

Answer 1

I am not clear why you have had trouble.

 

The equation of motion in the vertical direction is

 

s(t) = s₀ + v₀t+ 4.9t² = 40 + 15t sin 30° - 4.9t²

 

s'(t) = 7.5 - 9.8t and the maximum height will be attained when s'(t) = 0, i.e. at t= 7.5/9.8 = .765306

 

The value of s(.765306) = 42.869899

 

You have not asked about lateral motion, but the equation of motion will have no gravity component and the initial velocity will be modified by the cos 30°.