You are standing on top of building 135 m tall. You throw ball upward with veloc

Question

You are standing on top of building 135 m tall. You throw ball upward with velocity of 22.0 m/s. At exact moment a friend throws a second ball upward from ground with velocity of 46.0 m/s these 2 balls then collide at some later time.
a. How long after these 2 balls are released will they collide?
b. Where will these two balls be when they collide?
c. What will be the velocity of each ball just as they collide?
d. What will be relative velocity between these two balls at the moment they collide?

Explanation / Answer

let the distance from the top to the point where balls collide be = x so;the distance from the bottom to the point where balls collide be =135- x let A be the ball thrown from top so; for A :--------------------------------------------- uA = -22m/sec acceleration = a = 10 m/sec^2 displacement = x VA = uA + at-----------------------------------------------------------(n) so; x = (uA)*t + 1/2 at^2 = -22t + 5t^2 x = -22t + 5t^2 --------------------------------------(1) let B be the ball thrown from top so; for B :--------------------------------------------- uB = 46m/sec acceleration = a = -10 m/sec^2 VB = uB + at-----------------------------------------------------------(m) displacement = 135-x so; 135-x = (uB)*t + 1/2 at^2 = 46t - 5t^2 135-x = 46t - 5t^2--------------------------------------(2) adding (1) and (2) we get:--- 135 = 24 t so; t = 5.6 sec ---------------------------------------------------Ans putting the value of t in equ(1) we get x --------------------------------Ans putting the value of t in equ (m) and (n) we get the velocity of balls just before they collide relative velocity of ball A with respect to B = |VA|+|VB| --------------Ans |VA| means that only the magnitude of VA similarly for |VB|

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