Question

A blue ball is thrown upward with an initial speed of 21.8 m/s, from a height of 0.9 meters above the ground. 2.7 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 10.4 m/s from a height of 26.6 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.

Answer 1

To go through the basket, the ball needs to be thrown with an initial speed of approximately 25.2 m/s.

Step-by-step explanation:

The initial speed at which the ball needs to be thrown in order to go through the basket can be determined using the equations of projectile motion. We can break the initial velocity of the ball into its horizontal and vertical components. The horizontal component will determine the distance the ball will travel, and the vertical component will determine the height at which the ball will pass through the basket.

Using the given information, including the angle at which the ball is released and the height of the basket, we can calculate the initial speed of the ball using trigonometry and projectile motion equations. The initial speed required for the ball to go through the basket is the magnitude of the total initial velocity, which is the vector sum of the horizontal and vertical components.

The initial speed required for the ball to go through the basket is approximately 25.2 m/s.

Answer 2

I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:

When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.

1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²

For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.

2. For this, you equate the y values of both balls:

y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t,
t = 2.25 seconds

Thus, the two balls would be at the same height after 2.25 seconds.