A red ball is thrown down with an initial speed of 1 m/s from a height of 27 meters above the ground.

Question

A red ball is thrown down with an initial speed of 1 m/s from a height of 27 meters above the ground. Then, 0.7 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24.7 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.

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

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

3)What is the maximum height the blue ball reaches?

4)What is the height of the blue ball 1.8 seconds after the red ball is thrown?

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

Answer 1

1) Let’s start out with trajectory equations for the red ball. To find the speed when it hits the ground we can use the trajectory equation describing the final velocity squared = the initial velocity squared +2 times the acceleration of gravity times ▲X. Solving for final velocity using 1 m/s squared for the initial velocity and -9.81 for the acceleration of gravity, I got final velocity as it hits the ground to be 23.04 m/s

2) to find the time it takes to hit the ground I used the trajectory equation of ▲ X (27) = (final velocity plus initial velocity)/2,  rendering 2.25 seconds.

3) to find the maximum height of the blue ball, I used final velocity squared (0) at the max point = initial velocity squared +2 times (-9.81) ▲ X. Solving for ▲ X,  I got 31.1 but we need to add this to initial height from which the ball was launched at 0.8 meters. Here we get 32 m for the maximum. At this point you can actually check your work by using the other trajectory equation of the final velocity = initial velocity plus acceleration times time. Solving for t, I got 2.52 seconds. Putting that into the ▲ X = final velocity of 0+ initial velocity of 24.7÷2x 2.52, rendering 31.1 meters. Then we have to add the 0.8 initial condition rendering 32 m as a check.

4) now we need to find the height of the blue ball 1.8 seconds after the red ball is launched. Since the blue ball is delayed by .7 seconds compared to the red ball we will use the time in air to be 1.8 -.7 = 1.1 seconds after blue ball is launched. So using the equation ▲X = initial velocity times time +1/2 gravity times time squared,  using 1.1 seconds renders a ▲ X of 20.8 m. Again here we have to add 0.8 meters as the initial condition for the blue ball. This renders the height of the blue ball to be 21.6 meters.

5) now here comes the really hard part. Try to express the trajectory of the red ball in terms of its initial condition of 27 m and try to express the trajectory of the blue ball .7 seconds after the red ball is launched. Here are the equations I used: for the red ball

 (27-▲x)= initial velocity (1)t+1/2gt² , solving for▲x=-4.9t²-t+27.

Now for the blue ball, let’s express its trajectory using the equation:

(▲x-.8)= initial velocity (24.7)t-4.9t², now we have to delay the time factor by 0.7 seconds so the equation becomes for the blue ball:

▲x=24.7(t-0.7)-4.9(t-0.7)²⁾+0.8. Setting the ▲ X for the red ball equal to the ▲ X for the blue ball using a graphic calculator,  the time came out to be approximately 1.4 seconds meeting at a height of 15.69 meters.

Please be sure to check the reasonability of step 5 above.