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Projectile homework problem

  • Thread starter luvsk8ing
  • Start date
10
0
Hello.

1. 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

2. 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)


3. 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.:biggrin:
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

Integral
Staff Emeritus
Science Advisor
Gold Member
7,184
55
Since you do not know how high the ball went, you need to break the problem into 2 pieces. One piece is when the ball reaches the height of the roof (20m) reaches its highest point and returns to the roof. The second piece is the flight from the ground to roof height. You need to find how long the ball spends on the trip up to the roof. Once you get that time you know how long the rest of the flight took.

See if you can use that to finish the problem.
 
andrevdh
Homework Helper
2,126
116
..... then, Vo:
Vo=(24.6 m/s) / (sin 60°)=28.4 m/s ....

How do you now that the launching angle is sixty degrees?

(a) Using

[tex]v_y = v_{y0} - gt[/tex]

one get that

[tex]v_{y0} = v_y + gt[/tex]

substitute that into

[tex]y = v_{y0}t - 0.5gt^2[/tex]

to get

[tex]y = v_yt + 0.5gt^2[/tex]

using the data when it lands on the roof you can solve for the y-velocity component, which enables you to get the x-velocity component since

[tex]\frac{v_{y\ roof}}{v_x} = \tan(60^o)[/tex]

Do you have the entry page (contents/chapters) address for the Wiley notes?
 
Last edited:

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