Question

A red ball is thrown down with an initial speed of 1 m/s from a height of 26 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 23.5 m/s, from a height of 1 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s².

What is the speed of the red ball right before it hits the ground?

How long does it take the red ball to reach the ground?

What is the maximum height the blue ball reaches?

What is the height of the blue ball 1.9 seconds after the red ball is thrown?

How long after the red ball is thrown are the two balls in the air at the same height?

Answer 1

Speed of the red ball right before it hits the ground
Take v as the velocity with which the ball hits the ground,
u be the initial velocity,here u = 1 m/s
s be the distance it travelled, s = 26 m (height from which it is thrown),
a be the acceleration which is 9.81 m/s².
Substituting these in the equation
v² - u² = 2as,
v = sqrt (u² + 2as)
? = sqrt (1² + 2 x 9.81 x 26)
  = sqrt ( 511.12)
  = 22.6 m/s.

Time taken by the red ball to reach the ground
Let v be the final velocity, v = 22.6 m/s (previous question),
u be the initial velocity, u = 1 m/s, a be the acceleration, a = 9.81 m/s².
Substituting these values in the equation
v = u + at,
t = (v - u)/a,
= (22.6 - 1)/9.81 = 2.2 s

The maximum height the blue ball reaches
At the maximum height, velocity will be zero, v = 0
Initial velocity, u = 23.5 m/s, acceleration a = -9.8 m/s²? (here negative because acceleration is in the opposite direction of motion) ,
Take s be the maximum height,
Substituting these values in the equation,
v² - u² = 2as,
s = (v² - u²)/2a
= (0 - 23.5²) / 2 x -9.8 m/s².
  = 28.14 m
Since it is thrown 1 m above the ground, maximum height, h = 1 + 28.14 = 29.14 m.

Height of the blue ball 1.9 seconds after the red ball is thrown
The blue ball is thrown after 0.6 s the red ball is thrown,
So when the red ball has travelled 1.9 s, the blue ball will travel only 1.9 - 0.6 = 1.3 s
Let s be the height it reached after 1.3 s,
u be the initial velocity which it is thrown, u = 23.5 m/s,
a be the acceleration, here a = -9.8 m/s² (here negative because acceleration is in the opposite direction of motion).
Substituting these values in the equation,
s = u t + 0.5 a t²,
s = 23.5 x 1.3 + (0.5 x -9.8 x 1.3²),
= 30.55 - 8.29
  = 22.26 m

Time when the two ball reach the same height
Let t be the time for the red ball to travel s m, It is thrown with a velocity u = 1 m/s, So the equation of motion of the red ball will be,
s = u t + 0.5 a t²,   
s = 1 x t + 0.5 x 9.81 x t² ......(1)

When the red ball travelled t seconds, the blue ball will travel only (t - 0.6) s
It is thrown up with a velocity v = 23.5 m/s, it is feeling the acceleration, a = - 9.8 m/s²
For these two balls to reach the same height, when red ball reachs s m, the blue ball should be at (25 -s) m ( The distance is 25 m since it is thrown 1 m above the ground)
So the equation of motion of the blue ball is
25 - s = v x ( t - 0.6) - 0.5 a (t - 0.6)².
25 - s = 23.5 x (t - 0.6) - 0.5 x 9.81 x (t - 0.6)² ......(2)
(1) + (2) gives,
25 = t +0.5 gt² +23.5 t - 23.5 x 0.6 - [0.5 g (t² -1.2 + 0.36)],
Simplifying this will get,
t (30.386) = 37.3342
so, t = 1.228 s