Solving Ten-Pin Bowling Physics with Proportional Signs

In summary: So the mass becomes m(1+0.2)=1.2m?In summary, the conversation is about a novice ten-pin bowler struggling with calculating the components, maximum height, time flight, and distance traveled by a lofted ball. They also discuss how to calculate these values if the ball was 20% heavier, focusing on representing the decrease in acceleration mathematically and calculating the new initial velocity and time to reach maximum height.
  • #1
moobs93
1
0
i know this is really simple, but I'm not very good at working with proportional signs..

a novice ten-pin bowler, slightly "lofts" the ball (releases it too late and lifts it into the air so that it crashes to the ground. If the ball was released with a velocity of 3.5 m/s at an angle of 15 degrees above the horizontal, calculate:

a) the x and y components
b) the maximum height achieved by the ball
c) the time flight of the ball
d) how far down the lane the ball lands, measured from the release point
e) all of the above again, if the ball was 20% heavier

so i got the (a)-(d) taking the original mass as m. and started (e) by assuming the second mass would be (m+0.2m) and that the force applied by the bowler on the ball, would remain constant so the acceleration would change in proportion to the mass change of an additional 20%

my first question is; how do you represent the decrease in acceleration mathematically?
using F=ma, where m is now m(1+0.2)
my belly-flop attempt:
F=ma
a=K/m (where K is the constant force)
so mass is inversely proportional to the acceleration
does that mean the new acceleration is a(1-0.2) or am i completely off it?

and my second question is, how do you calculate the new initial velocity from the new acceleration? or rather, how do you calculate the time it takes to reach the maximum height (highest vertical displacement) using the new value for accleration? which i can then use to solve for the new initial velocity

once i have that i can finish the question myself
thanks heaps
 
Physics news on Phys.org
  • #2
moobs93 said:
i know this is really simple, but I'm not very good at working with proportional signs..

a novice ten-pin bowler, slightly "lofts" the ball (releases it too late and lifts it into the air so that it crashes to the ground. If the ball was released with a velocity of 3.5 m/s at an angle of 15 degrees above the horizontal, calculate:

a) the x and y components
b) the maximum height achieved by the ball
c) the time flight of the ball
d) how far down the lane the ball lands, measured from the release point
e) all of the above again, if the ball was 20% heavier

so i got the (a)-(d) taking the original mass as m. and started (e) by assuming the second mass would be (m+0.2m) and that the force applied by the bowler on the ball, would remain constant so the acceleration would change in proportion to the mass change of an additional 20%

my first question is; how do you represent the decrease in acceleration mathematically?
using F=ma, where m is now m(1+0.2)
my belly-flop attempt:
F=ma
a=K/m (where K is the constant force)
so mass is inversely proportional to the acceleration
does that mean the new acceleration is a(1-0.2) or am i completely off it?

and my second question is, how do you calculate the new initial velocity from the new acceleration? or rather, how do you calculate the time it takes to reach the maximum height (highest vertical displacement) using the new value for accleration? which i can then use to solve for the new initial velocity

once i have that i can finish the question myself
thanks heaps

Welcome to physics forums.

I interpret the question to mean the same given initial conditions apply (3.5 m/s at 15° to the horizontal), and the ball is 20% heavier.
 

Related to Solving Ten-Pin Bowling Physics with Proportional Signs

What is the concept behind "Solving Ten-Pin Bowling Physics with Proportional Signs"?

The concept behind "Solving Ten-Pin Bowling Physics with Proportional Signs" is to use mathematical equations and proportional signs to analyze and understand the physics behind the sport of ten-pin bowling. This can help bowlers improve their technique and performance on the lanes.

How do proportional signs play a role in solving ten-pin bowling physics?

Proportional signs, such as the equal sign (=) and the greater than/less than signs (> and <), are used in mathematical equations to show the relationship between different variables, such as speed, angle, and force, in the physics of ten-pin bowling. By understanding these relationships, bowlers can make adjustments to their approach and release to improve their accuracy and power.

What type of physics concepts are involved in ten-pin bowling?

Ten-pin bowling involves various physics concepts, including force, velocity, acceleration, friction, and energy. These concepts all play a role in how a bowling ball moves down the lane and interacts with the pins.

How can understanding the physics of ten-pin bowling help improve performance?

Understanding the physics of ten-pin bowling can help bowlers improve their performance in several ways. By understanding the relationship between speed and angle, for example, bowlers can make adjustments to their approach and release to hit the pocket more consistently. Additionally, understanding the concept of energy transfer can help bowlers maximize the power of their throw and increase pin carry.

Are there any limitations to using proportional signs to solve ten-pin bowling physics?

While using proportional signs can help us understand the fundamentals of ten-pin bowling physics, there are limitations. For example, there are many variables at play in a bowling shot, such as lane conditions and the unique characteristics of each bowler's style. These factors can make it difficult to create a one-size-fits-all equation for solving bowling physics. Additionally, the use of proportional signs does not account for the mental and strategic aspects of the sport.

Similar threads

  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
401
  • Introductory Physics Homework Help
Replies
10
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
4K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
24
Views
2K
Back
Top