- #1
luvsk8ing
- 10
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Hello.
There is a link to the problem and its picture here: http://media.wiley.com/product_data/excerpt/19/04717580/0471758019-1.pdf
It is # 46.
Here is the problem as well:
In Fig. 4-44, a ball is thrown up onto a roof, landing 4.00 s later at height h=20.0m above the release level. The ball’s path just before landing is angled at θ = 60.0° with the roof. (a) Find the horizontal distance d it travels. (See the hint to Problem 41.) What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball’s initial velocity?
Given/Known:
t=4.00s
h=20.0m
θ=60.0°
Yo=0
Y=20.0m
ΔY=h
I used these:
Voy=(Y-Yo + 0.5gt²)/t
Vo=Voy/(sin θ)
Vox=Vo (cos θ)
Δx=(Vo cos θ) t
magnitude= √[(Vox)²+(Voy)²]
θ=arctan (Y/X)
Solved for Voy, first:
Voy= [20 m + (0.5)(9.8 m/s²)(4 s)²]/ 4 s
=24.6 m/s
then, Vo:
Vo=(24.6 m/s) / (sin 60°)=28.4 m/s
Vox=(28.4 m/s) (cos 60°)= 14.2 m/s
Want:
a) Δx = (28.4 m/s)(cos 60°) (4 s)
= 56.8 m
b) magnitude (V)= √[(14.2 m/s)²+(24.6 m/s)²]= 28.4 m/s
c) θ= (tan ^-1)[(24.6 m/s)/(14.2 m/s)] = 60°
I'm worried my answers seem wrong. Any help would be greatly appreciated.
Homework Statement
There is a link to the problem and its picture here: http://media.wiley.com/product_data/excerpt/19/04717580/0471758019-1.pdf
It is # 46.
Here is the problem as well:
In Fig. 4-44, a ball is thrown up onto a roof, landing 4.00 s later at height h=20.0m above the release level. The ball’s path just before landing is angled at θ = 60.0° with the roof. (a) Find the horizontal distance d it travels. (See the hint to Problem 41.) What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball’s initial velocity?
Given/Known:
t=4.00s
h=20.0m
θ=60.0°
Yo=0
Y=20.0m
ΔY=h
Homework Equations
I used these:
Voy=(Y-Yo + 0.5gt²)/t
Vo=Voy/(sin θ)
Vox=Vo (cos θ)
Δx=(Vo cos θ) t
magnitude= √[(Vox)²+(Voy)²]
θ=arctan (Y/X)
The Attempt at a Solution
Solved for Voy, first:
Voy= [20 m + (0.5)(9.8 m/s²)(4 s)²]/ 4 s
=24.6 m/s
then, Vo:
Vo=(24.6 m/s) / (sin 60°)=28.4 m/s
Vox=(28.4 m/s) (cos 60°)= 14.2 m/s
Want:
a) Δx = (28.4 m/s)(cos 60°) (4 s)
= 56.8 m
b) magnitude (V)= √[(14.2 m/s)²+(24.6 m/s)²]= 28.4 m/s
c) θ= (tan ^-1)[(24.6 m/s)/(14.2 m/s)] = 60°
I'm worried my answers seem wrong. Any help would be greatly appreciated.