Projectile Motion in One Dimension

[razrsharp67]

Homework Statement

When a ball was thrown and caught at the same height from which it was thrown, it was measured to have traveled 12 meters in 1.3 seconds. What was the launch velocity?

The answer on the key says 11.2 meters per second.

Homework Equations

Yfinal = Yinit + Vt + 1/2 a t^2

The Attempt at a Solution

If it goes 12 meters in 1.3 seconds, then it hit its peak in .65 seconds at 6 meters.

So Yfinal = 6 and Yinit = 0 and t = .65 and a = -9.81

Plugging that all in and solving for V I get 12.419 m/s. But my teacher on the key says the answer is 11.2 m/s. Where am I going wrong?

 

[Ace.]

v_f² = v_i² + 2ad

looking at first half of the balls journey.

v_f = 0 m/s
a = -9.8 m/s² [up]
d = 6 m

Solve for v_i

v_i = √(v_f² - 2ad)

= √(-2ad)
=√(-2(-9.8)(6))
=10.84 m/s [up]getting a different answer :uhh:. I did not use time in my calculation though.

 

[razrsharp67]

I tried that method too, but that answer also wasn't an answer choice.

 

[haruspex]

When a ball was thrown and caught at the same height from which it was thrown, it was measured to have traveled 12 meters in 1.3 seconds. What was the launch velocity?

It's not as clear as it might be, but I suspect the question means that it traveled 12m horizontally.

 

[PeterO]

Homework Statement

When a ball was thrown and caught at the same height from which it was thrown, it was measured to have traveled 12 meters in 1.3 seconds. What was the launch velocity?

The answer on the key says 11.2 meters per second.

Homework Equations

Yfinal = Yinit + Vt + 1/2 a t^2

The Attempt at a Solution

If it goes 12 meters in 1.3 seconds, then it hit its peak in .65 seconds at 6 meters.

So Yfinal = 6 and Yinit = 0 and t = .65 and a = -9.81

Plugging that all in and solving for V I get 12.419 m/s. But my teacher on the key says the answer is 11.2 m/s. Where am I going wrong?

The Title says this is Projectile Motion, so this projectile travels up for 0.65 sec, then comes down for 0.65. What speed does an object reach in 0.65 seconds [a little less than 6.5 m/s I suggest]
Horizontally speaking - the projectile has traveled 12m in 1.3 seconds; that means a horizontal component of a little less than 10.

Lets imagine those values were 6 and 10. Using Pythagoras gives v = √136 which is about 11.6 [note 136 = 6² + 10²
No doubt if you use the correct values you will get the 11.2 you are after.