Bowling Ball Thrown off Building-Check my work/solutions pls.

• afg_91320
In summary: Keep up the good work.In summary, a ball rolls down a roof at an angle of 30 degrees to the horizontal with a speed of 5.00 m/s. The distance to the ground from that point is 7.00 m. Using the formula y = Vo*t + .5*a*t^2, the time in the air is found to be 0.97 seconds. The ball lands 6 meters from the base of the house. To find its speed before landing, the initial velocity is broken into its horizontal and vertical components, and then combined to get the final velocity.
afg_91320

Homework Statement

A ball rolls down a roof that makes an angle of 30 degrees to the horizontal. It rolls
off the edge woth speed of 5.00 m/s. The distance to the ground from that point is 7.00 m.
a) How long is the ball in the air for?
b) How far from the base of the house does it land?
c) What is its speed before landing?

Homework Equations

$$\vartheta$$ = 30 degrees
speed = 5.00 m/s
distance from ground to point ball being dropped (roof) = 7.00 m

The Attempt at a Solution

a) time in the air:
y = -7.00m
vx0 =5.00 m/s
ax = 0 <- drop so acceleration is 0..correct?
vy0 = 0
ay = -g <---gravity
x0 and y0 = 0

y = - 1/2 gt2
t = sqrt [(2y)/(-g) = sqrt [(2)(-7.00)/(-9.80) = 1.2 s <--ball in the air for

b) far from the base of the house:
x = vx0t = (5.00)(1.2) = 6 m<---- distance from base of the house

c) i don't know how to find the speed-do i need the angle in this equation?

I don't think part a is correct.
In part (a), you are only concerned with the vertical motion so the distance is 7 m as you have, but the initial velocity is only 5*sin(30). The initial velocity is not zero so you must use the whole formula for distance: y = Vo*t + .5*a*t^2.

but then i get -2.88 <---since a was -g (-9.80)

initial velocity was: 2.5
and then y came out to be -4.06.

is that correct? is my base distance correct??

I don't get 2.88. Isn't y = 7?
What formula are you using?

this is what i used:
y = Vo*t + .5*a*t^2.

^unless you are talking about initial y!

y = Vo*t + .5*a*t^2.
7 = ?

7 = (5 sin 30)(1.2) + .5(-9.80)(1.2)^2
v0 = 5 sin 30
t = 1.2 s
a = -g = -9.80
7 = 3 + (-7.056)
7 = -4.056 <---that doesn't make sense, am i suppose to solve for a variable?

"7 = (5 sin 30)(1.2) + .5(-9.80)(1.2)^2"
Okay - now I can see what you are doing!
First, you don't know what the time is, so you shouldn't be putting in 1.2 for t.
In fact, we are trying to find the time, so it must appear as a t.
Second, down is positive in this problem, so drop the minus sign on the 9.8.

Then you are ready to solve the formula for time. Looks like it is a quadratic!

oohhh!

i thought i solved for time correctly-i got 1.2s for t.
so it is:
7 = (5 sin 30)(t) + .5(9.80)(t)^2
7 = 2.5t + 4.9t^2
0 = 4.9t^2 + 2.5t -7
and time should be in the positive correct?

now how do i find the speed??

Yes, looks correct. What time did you find?
Speed should be done in two parts - vertical and horizontal.
Horizontal is easy - no acceleration.
Vertically not bad when you know the initial speed and the time of fall.

im using the quad formula which i assume is the way to go.
but my answers are coming in the neg. I am trying to rework it.

what do you mean by vertical and horizontal? how do i find that? (sorry its my first time takin physics) you mentioned it in my other prblm as well so i want to make sure I am doing it rite.

by the way-thanks for helpin me with my prblms! it means a lot! :D

ok so here is my solution for t:
(-2.5 + 12) / 9.8 = 0.97 s

The quadratic formula gives one negative and one positive answer. Ignore the negative - something that happened before the problem started!

You have 5 m/s at 30 degrees below horizontal. Make a triangle around that with a horizontal side and a vertical side. The vertical side has length 5 sin(30). The horizontal side has length 5*cos(30). That is your horizontal velocity.

Use V = Vo + at to get the vertical velocity at the time it reaches the ground.

Finally, you must combine the horizontal and final vertical velocities using a triangle again. This time you must calculate the hypotenuse to get the combined velocity.

Most welcome - thanks for using PF.

1. How does the height of the building affect the speed of the bowling ball?

The height of the building does not affect the speed of the bowling ball. The speed of the ball is determined by its initial velocity and the force of gravity acting on it. The height of the building only affects the time it takes for the ball to hit the ground.

2. Does the weight of the bowling ball make a difference in its trajectory?

Yes, the weight of the bowling ball does make a difference in its trajectory. Heavier objects have a greater resistance to changes in motion, so a heavier bowling ball will have a slightly different trajectory than a lighter one thrown with the same initial velocity.

3. How does air resistance affect the motion of the bowling ball?

Air resistance or drag does have an impact on the motion of the bowling ball. As the ball falls, it will experience air resistance which will slow it down. However, this effect is negligible for a heavy object like a bowling ball compared to lighter objects.

4. Can the angle at which the ball is thrown affect its speed?

Yes, the angle at which the ball is thrown can affect its speed. When thrown at an angle, the initial velocity of the ball is divided into horizontal and vertical components. The vertical component is affected by the force of gravity, while the horizontal component remains constant. This can result in a slower or faster speed depending on the angle of the throw.

5. Is there a maximum speed that a bowling ball can reach when thrown off a building?

Yes, there is a maximum speed that a bowling ball can reach when thrown off a building. This is known as terminal velocity and is dependent on the weight and surface area of the ball. Once the force of air resistance equals the force of gravity, the ball will stop accelerating and reach a constant speed.

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