Physics projectile question.

Question

You’re 6.00 m from one wall of a house. You want to toss a ball to your friend who is 6.00 m from the opposite wall. The throw and catch both occur at 1.00 m above the ground.

 

Here is a link to the diagram that was given in the question: https://answers.yahoo.com/question/index?qid=20150509165726AA1yK5z

a. What is Vy0 if it just barely clears the roof?

b. How long is it in the air when it just clears the roof?

c. What minimum launch speed that will allow the ball to clear the roof?

d. At what angle should you toss the ball?

 

 

 

Answer 1

From the diagram we can tell that the two people are 18 meters apart and that the roof is 9 meters high.   Without that, the problem cannot be solved.

 

The ball starts out at one foot high rises to 9 meters and then falls again to one foot high.  All while traveling 18 meters horizontally.   Looking at the last half of the flight, the ball falls from 9 meter high to one meter.  We can solve for the required for an 8 meter fall using d=1/2(g)t².   t = 1.28s.  The rise from 1 m to 9 m took the same amount of time, and the vertical velocity at the peak is zero.

 

V_f = V_i - gt => 0 = V_i -9.8(1.28) => V_i = 12.5 m/s

 

b. is already found to be 1.28s

 

The remaining answers are trivially easy.