Homework Problem: How to solve a projectile problem with given information?

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The homework problem involves calculating the horizontal distance a ball travels after being thrown onto a roof, landing 4 seconds later at a height of 20 meters with a landing angle of 60 degrees. The initial vertical velocity was calculated to be 24.6 m/s, leading to an initial velocity of 28.4 m/s. The horizontal distance traveled was determined to be 56.8 meters, and the magnitude of the initial velocity was confirmed as 28.4 m/s with an angle of 60 degrees. The discussion emphasizes breaking the problem into two parts: the ascent to the roof height and the descent. Additional clarification on the launch angle and the use of kinematic equations was provided to assist in solving the problem.
luvsk8ing
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Hello.

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.:biggrin:
 
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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.
 
... 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

v_y = v_{y0} - gt

one get that

v_{y0} = v_y + gt

substitute that into

y = v_{y0}t - 0.5gt^2

to get

y = v_yt + 0.5gt^2

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

\frac{v_{y\ roof}}{v_x} = \tan(60^o)

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