# Thread: Kinematics Problem [Related to Velocity and displacement]

1. ## Kinematics Problem [Related to Velocity and displacement]

I have the following problem...

A person throws a ball upward into the air with an initial velocity of 15.0m/s. Calculate: how high it goes, how long is in the air before it comes back to his hand, the velocity of the ball when it returns to the thrower's hand, and the time when the ball passes a point 8.0m above the person's hand.

With so little given [only the initial velocity is given, and I know that gravity plays a part, but I am still stumped], I'm not sure how to proceed with this. Help is appreciated.

Thank you!

2. Originally Posted by nathan02079
I have the following problem...

A person throws a ball upward into the air with an initial velocity of 15.0m/s. Calculate: how high it goes, how long is in the air before it comes back to his hand, the velocity of the ball when it returns to the thrower's hand, and the time when the ball passes a point 8.0m above the person's hand.

With so little given [only the initial velocity is given, and I know that gravity plays a part, but I am still stumped], I'm not sure how to proceed with this. Help is appreciated.

Thank you!
what kinematics equations are you familiar with?

3. All the basic ones...

$\displaystyle a={{v_f-v_0}\over{t}}$

$\displaystyle d={{v_f+v_0}\over{2}}t$

$\displaystyle d=v_0t+{1\over2}at^2$

$\displaystyle {v_f}^2={v_0}^2+2ad$

$\displaystyle a$ = acceleration
$\displaystyle v_f$ = final velocity
$\displaystyle v_0$ = initial velocity
$\displaystyle d$ = displacement
$\displaystyle t$ = time

4. you'll need at a minimum these two (modified for vertical motion in a uniform gravitational field) to solve your problem ...

$\displaystyle \Delta y = v_o t - \frac{1}{2}gt^2$

$\displaystyle v_f = v_o - gt$

5. Just a question--what is the $\displaystyle y$ variable in your case? The displacement?

6. Originally Posted by nathan02079
Just a question--what is the $\displaystyle y$ variable in your case? The displacement?
$\displaystyle \Delta y$ = the vertical displacement

7. Okay, I'm just a *bit* confused, but I'll try. This was actually a bonus question that I'm trying to solve. :P

So If I want to find the time first, do I replace the second equation you gave me with the first equation I listed?

$\displaystyle g={{(v_o - gt)-v_o}\over{t}}$

8. (1) how high it goes ...

at the top of its trajectory, $\displaystyle v = 0$

$\displaystyle (v_f)^2 = (v_o)^2 - 2g(\Delta y)$

$\displaystyle 0 = 15^2 - 2g(\Delta y)$

solve for $\displaystyle \Delta y$

(2) how long the ball is in the air ...

when the ball returns to its starting position, $\displaystyle \Delta y = 0$

$\displaystyle \Delta y = v_o t - \frac{1}{2}gt^2$

$\displaystyle 0 = 15t - \frac{1}{2}gt^2$

solve for $\displaystyle t$

(3) velocity of the ball when it returns ...

once you have $\displaystyle t$ from part (2)

$\displaystyle v_f = v_o - gt$

$\displaystyle v_f = 15 - gt$

evaluate $\displaystyle v_f$

(4) time when the ball is 8.0 m above ...

$\displaystyle \Delta y = v_o t - \frac{1}{2}gt^2$

$\displaystyle 8 = 15t - \frac{1}{2}gt^2$

solve the quadratic for t ... you'll get two solutions, one time on the way up and one time on the way down.

9. Wow, thanks! Now I feel so stupid.