3. Two kids are playing an unconventional game of catch One kid is standing on t

Question

3. Two kids are playing an unconventional game of catch One kid is standing on the ground and throwing a ball upward, while the other kid is standing on the second floor of a house and throwing a ball downward. Alex throws a ball upward from a height of 0.85 m with a speed of 8.1 m/s. Bruno throws a ball downward from a height of 4.1 m above the ground with a speed of 8.1 m/s. They each throw at the same instant. a. Find the time it takes for the two balls to cross paths after thev are thrown paths. high enough for Bruno to catch it? threw it from, would it be traveling faster, slower, or b. Find the height above the ground where the balls cross c. Will Alex's ball travel to a maximum height that is d. If Alex caught his own ball at the same height he the same speed as Bruno's ball would at that height? Explain your answer using words or equations.

Explanation / Answer

Alex throws the ball upward from a height 0.85 m from the ground

with initial velocity u1 = 8.1 m/s

and Bruno drops the ball from a height 4.1 m from the ground with initila velocity is u2 = 8.1 m/s

let the vertical distance covered by ball2 (Bruno) in time t sec is  

(4.1-h) = 8.1*t+0.5*9.8*t^2

h = 4.1-8.1*t+ -0.5*9.8*t^2 ------(1)

and in that time Alex ball 1 moves to a height so that  

h = 8.1*t -0.5*9.8*t^2 --------(2)

equating both  

4.1-8.1*t+ -0.5*9.8*t^2 = 8.1*t -0.5*9.8*t^2

solving for t , t = 0.253 s

substituting the value of t in eq(2)

h = 8.1*0.253 -0.5*9.8*0.253^2 m  

h = 1.7356559 m = 1.74 m

a) time taken for the two balls to cross paths after they are thrown is t= 0.253 s

b) the balls can cross paths is at h = 1.74 m from the ground

c) maximum height reached by the Alex's ball is h1 = u^2/2g = 8.1^2/(2*9.8) m = 3.35 m

No Burno will not catch the ball1

d)

from equations of motions the speed of the ball1 at the same point where it was thrown is same speed

from conservation of energy mgh = 0.5*mv^2

v = sqrt(2*g*h) = sqrt(2*9.8*0.89) = 4.18 m/s slower

Burno ball2 speed at the height 0.85 m from the ground is  

mgh = 0.5*m*v^2

v = sqrt(2*g*h) = sqrt(2*9.8*(4.1-0.85)) m/s

===> v = 7.98 m/s

so the speed of the ball1 at 0.85 m from the ground is slower than its initial velocity and also slower than ball2

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