How Fast Must a Quarterback Throw to Achieve a 171 Meter Pass?

Click For Summary
SUMMARY

The discussion centers on calculating the speed required for a quarterback to throw a football a horizontal distance of 171 meters at an angle of 26.2 degrees. The relevant formula for this projectile motion is R = v² * sin(2θ) / g, where R is the range, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (9.80 m/s²). The user expresses difficulty in solving the problem due to a perceived lack of information but is guided to use the established formula to find the necessary throwing speed.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions, specifically sine
  • Knowledge of basic physics equations related to motion
  • Ability to manipulate algebraic equations to isolate variables
NEXT STEPS
  • Study the derivation of the projectile motion range formula R = v² * sin(2θ) / g
  • Learn how to apply trigonometric identities in physics problems
  • Explore the effects of launch angle on projectile distance
  • Investigate real-world applications of projectile motion in sports
USEFUL FOR

Students studying physics, coaches analyzing quarterback performance, and anyone interested in the mechanics of projectile motion in sports contexts.

Phoenix838
Messages
2
Reaction score
0

Homework Statement



A quarterback claims that he can throw the football a horizontal distance of 171 m. Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 26.2 ° above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.

Homework Equations



vx = v0x + axt
vy = v0y + ayt
x = 1/2(v0x + vx)t
y = 1/2(v0y + vy)t
x=v0xt + 1/2axt2
y=v0yt + 1/2ayt2
vx2 = v0x2 + 2axx
vy2 = v0y2 + 2ayy



The Attempt at a Solution



The difficulty that I'm having is it seems like I don't have enough information to solve the problem. I've got the horizontal distance (x) as 171 meters, the horizontal acceleration (ax) as 0 m/s2, the vertical distance (y) as 0 meters and the vertical acceleration (ay) as -9.80 m/s2.

I know all I need is one more variable in order to solve this problem, but I just can't seem to figure out what I'm missing given the information I've got.
 
Physics news on Phys.org
Using the relevant equations, you can show that the range of the projectile is given by
R = v^2*sin(2θ) /g. Using this formula find v.
 
rl.bhat said:
Using the relevant equations, you can show that the range of the projectile is given by
R = v^2*sin(2θ) /g. Using this formula find v.

Would you mind explaining how you came up with that formula?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
1
Views
2K
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 17 ·
Replies
17
Views
36K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
5
Views
6K