Question

A red ball is thrown down with an initial speed of 1.3 m/s from a height of 27 meters above the ground. Then, 0.5 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24 m/s, from a height of 0.8 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?

(Is it just jus the velocity equestion?)

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

Maybe I am not doing something right, but this is what I have: 27 - 1.3t - 4.9t^2 = -4.9t^2 + 28.9t-12.425
with -4.9t^2 + 28.9t-12.425 being derived from .8 + 24(t-.5)-4.9(t-.5)^2 the only thing I can thing of would be to maybe to have it as 8 + 24(t+.5)-4.9(t+.5)^2 ?

Is the acceleration postive or negative and is speeding up or slowing down?

Answer 1

For rhe red ball:

Let t_rbe be time taken by the red ball to reach the ground. Then,

s_r = u_rt_r + 0.5gt_r²

=> 27 = 1.3t_r + (0.5*9.81)t_r²

=> t_r² + 0.265t_r - 5.505 = 0

=> t_r = [-0.265 + (0.265² + 4*5.505)^1/2]/2 = 2.218 s

Scalar acceleration is positive. Red ball is speeding in downward direction. Vector acceleration has a magnitude of 9.81 m/s² and is directed downwards.

The blue ball rises for t = 24/9.81 = 2.45 s > 2.218 s. Hence, the blue ball will be at the same height as the red ball while it's going up.

Let T be the time after the red ball is thrown at which the two balls are at the same height.

Then,

27 = (distance travelled by red ball in time T) + (distance tavelled by blue ball in time [T - 0.5]) + 0.8

=> 27 = [1.3T + (0.5*9.81)T²] + [24(T - 0.5) - (0.5*9.81)(T - 0.5)²] + 0.8

=> 27 = (1.3T + 4.905T²) + [24T -12 - 4.905(T² + 0.25 - T)] + 0.8

=> (24 + 1.3 + 4.905)T = 27 - 0.8 + 12 + 4.905*0.25

=> T = 1.305 s