Question

A red ball is thrown down with an initial speed of 1.1m/s from a height of 28 meters above the ground. Then 0.5 seconds after the red ball is thrown, a blue bar) is thrown upward with an initial speed of 24.6m/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 9.81m/s^2

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 2 seconds after the red bal is thrown
5. How long after the red ball is thrown are the two balls in the air at the same height?

Answer 1

Using kinematic equations for uniformly accelerated motion and considering gravity's acceleration, we can find the various projectile motion characteristics of the red and blue ball. These include the final velocity and time to ground for the red ball, the maximum height for the blue ball, the height of the blue ball after a specific time, and the time when both balls are at the same height.

Step-by-step explanation:

Solution for Two Balls in Projectile Motion

To solve these projectile motion questions involving a red ball thrown downward and a blue ball thrown upward, we use kinematic equations for uniformly accelerated motion. The acceleration due to gravity is a constant 9.81m/s2.

1. Speed of Red Ball Before Impact

Using kinematic equations, the final velocity (v) of the red ball can be found with the equation v = u + at, where u is the initial velocity (1.1m/s downwards), a is the acceleration due to gravity (9.81m/s2), and t is the time taken to reach the ground.

2. Time Taken for Red Ball to Reach the Ground

We calculate time (t) using the equation s = ut + 0.5at2, where s is the displacement (28 meters), u is the initial velocity, and a is the acceleration. By solving the quadratic equation, we find the time taken for the red ball to reach the ground.

3. Maximum Height Reached by Blue Ball

The maximum height (h) is found when the velocity at the highest point is zero. We use vf2 = vi2 + 2ah, where vf is the final velocity (0 m/s at the top), vi is the initial velocity (24.6 m/s upwards), and a is the acceleration due to gravity (-9.81 m/s2).

4. Height of Blue Ball After 2 Seconds

Since the red ball is thrown 0.5 seconds before the blue ball, we calculate the position of the blue ball 1.5 seconds after it is thrown. We use the previous kinematic equations for this calculation.

5. Time When Both Balls Are at the Same Height

We'll set the displacements from their respective starting points equal to each other and solve for the time using the kinematic equations.

Answer 2

1.
v2 = u2 + 2gh

v2 = 1.12 + 2 * 9.8 * 28 = 550.01

v = 23.45 m/s

The speed of the red ball right before it hits the ground = 23.45 m/s.

2.

v = u + gt

23.45 = 1.1 + 9.8 * t

t = 2.28 seconds

The time it take the red ball to reach the ground = 2.28 seconds,

3.

Height above the ground = 0.8 m

v2 = u2 - 2gh

0 = 25.32 - 2* 9.8* h

h = 32.66 m.

The maximum height the blue ball reaches = (h + 1) = (32.66 + 1) = 33.66 m.

4.

Time of travel of the blue ball = (1.9 - 0.5) = 1.4 seconds.

s = ut - 0.5 gt2

s = 25.3* 1.4 - 0.5 * 9.8 * 1.42= 25.816 m.

The height of the blue ball 1.9 seconds after the red ball is thrown = (s + 1) = (25.816 + 1) = 26.816 m.

5.

Let the time of travel of the red ball be t seconds.

So the time of travel of the blue ball = (t - 0.5) seconds.

Both the balls are at the same height :

28 - (s) = 1 + (h) ................{"s" & "h" are the displacements of the red & the blue ball respectively.}

28 - (ut + 0.5 gt2) = 1 + (ut - 0.5gt2)

28 - (1.1 t + 0.5 * 9.8 t2) = 1+ (25.3 (t-0.5) - 0.5*9.8*(t-0.5)2)

Now we have to solve the above equation to find the time after which both the balls are at the same height.

28 - 1.1t - 4.9t2 = 1 + 25.3t - 12.65 - 4.9t2 + 4.9 t - 1.225

(28 - 0.8 + 12.65 +1.225) = (25.3 + 4.9 + 1.1) * t

t = 41.075 / 31.3 = 1.3123 = 1.31 seconds (approx.)

The time after the red ball is thrown are the two balls in the air at the same height = 1.31 seconds.