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Greeley [361]
3 years ago
6

What is the definition of distance?

Physics
1 answer:
AnnyKZ [126]3 years ago
8 0
An amount of space beteeen 2 things or people
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Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the s
sveta [45]

Answer:

b

Explanation:

Given:

- The ball is fired at a upward initial speed v_yi = 2*v

- The ball in first experiment was fired at upward initial speed v_yi = v

- The ball in first experiment was as at position behind cart = x_1

Find:

How far behind the cart will the ball land, compared to the distance in the original experiment?

Solution:

- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:

                                      v_yf = v_yi + a*t

Where, a = -9.81 m/s^2 acceleration due to gravity

            v_y,f = 0 m/s max height for both cases:

For experiment 1 case:

                                     0 = v - 9.81*t_1

                                      t_1 = v / 9.81

For experiment 2 case:

                                     0 = 2*v - 9.81*t_2

                                      t_2 = 2*v / 9.81

The total time for the journey is twice that of t for both cases:

For experiment 1 case:

                                     T_1 = 2*t_1

                                     T_1 = 2*v / 9.81

For experiment 2 case:

                                     T_2 = 2*t_2

                                     T_2 = 4*v / 9.81

- Now use 2nd equation of motion in horizontal direction for both cases:

                                     x = v_xi*T

For experiment 1 case:

                                     x_1 = v_x1*T_1

                                    x_1 = v_x1*2*v / 9.81

For experiment 2 case:

                                     x_2 =  v_x2*T_2

                                    x_2 = v_x2*4*v / 9.81

- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.

- Hence; take ratio of two distances x_1 and x_2:

                        x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v

Simplify:

                        x_1 / x_2 = 2  

- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.      

                                   

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