Engineering Determine the flight time and heights of 2 balls thrown vertically

[tremain74]

Homework Statement

Child A throws a ball vertically up with a speed of 10 m/s from the top of a shed 3 meters high. Child B on the ground at the same instant throws a ball vertically up with a speed of 13.5 m/s. Determine the time at which the two balls will be at the same height above the ground. What is the height? The answers were t = 0.857 s and h = 15.17.

Relevant Equations

I used the equations 3 + (10*x) = 13.5*x.

To find the time of this problem, I used the equations 3 + (10 * x ) = 13.5 *x. 3 = 3.5*x. Therefore x was 0.857. To start off with child A was already three feet above ground and I got the answer to the time. However I couldn't find the height.

 

[phinds]

show what work you did do to try to get the height.

 

[tremain74]

The one formula that I thought about was the projectile motion formula using y = -4.9t^2 + vo*t*sinO. Sin O(sin of theta). However this does not solve my problem to find the correct height.

 

[phinds]

Since the balls were thrown vertically, why would you think there would be an angle involved. Do you have ANY experience with projectile motion? It certainly seems as though you do not, so I would suggest you go back and study the basics of projectile motion.

 

[neilparker62]

Surprisingly the OP's answer for the first part is correct albeit using variable x instead of t. This is because in problems like this the quadratic term(s) in the respective motion equations cancel out and one is left with exactly the equation solved by the OP.

The OP also wrote the correct equation to find the height - just needed to replace "vo" with 13.5 m/s. and the angle of projection is 90 degrees if the ball is thrown vertically upward.

The given answer for the height at which the balls meet (are in same horizontal line) seems incorrect - should be 13.5*0.857-4.9*(0.857)^2=7.971 m. See following graph.

https://www.desmos.com/calculator/snmbofcveh