Calculating Average Speed to Catch a Ball Thrown from a Building

[NDiggity]

Homework Statement

A guy on the top of a 25m building throws a ball upwards with a initial vel of 12 m/s. A guy 31 m from the building catches it at the bottom of the building. What must be the average speed of the guy to catch it at the bottom.

Homework Equations

Using the Kinematics equations, and using acceleration = -9.80 m/s^2 and v= 0, I find the time up to be 1.22s. Then I find the distance up as 7.35m. Then I find the time it takes to fall down by adding 7.35 to 25 m, using initial velo as 0, and acceleration. But I get a negative on one side in this equation, x = vint(t) + (1/2)(a)(t^2). What am i doing wrong?

I know once I get this time I add it to 1.22s. Then I believe I use v=d/t and plug in 31m for distance and the total time.

 

[alphysicist]

Hi NDiggity,

Homework Statement

A guy on the top of a 25m building throws a ball upwards with a initial vel of 12 m/s. A guy 31 m from the building catches it at the bottom of the building. What must be the average speed of the guy to catch it at the bottom.

Homework Equations

Using the Kinematics equations, and using acceleration = -9.80 m/s^2 and v= 0, I find the time up to be 1.22s. Then I find the distance up as 7.35m. Then I find the time it takes to fall down by adding 7.35 to 25 m, using initial velo as 0, and acceleration. But I get a negative on one side in this equation, x = vint(t) + (1/2)(a)(t^2). What am i doing wrong?

I assume you used a= -9.8 m/s^2 like you did before. But what number did you use for x in that equation? If the distance from the top to the ground is (7.35 + 25) = 32.35, then x is not 32.35. (Close, but not quite.) Remember that the x in that equation is displacement, not distance. What do you get?

 

[Dick]

You aren't doing anything wrong. Except that x in your formula is a displacement from the original position. With your sign conventions, that IS -(7.35+25)m. Now there's a negative on both sides. Just go ahead and solve it as you intended. Think about what the signs mean in terms of the problem and your coordinates.

 

[NDiggity]

Ahh, so since I was using up as positive, displacement would be -32.35 since the ball is going down. Thank you both so much!

 

[alphysicist]

An alternative way is that you could actually find the time in one step, from the initial throw to the ball hitting the ground (insted of using three equations). Displacement is -25, initial velocity is +12, so:

$$(-25)= (+12) t + \frac{1}{2}(-9.8) t^2$$

Of course you'll have to use the quadratic formula to find t doing it this way, and you'll keep the positive answer.