Question

using a powerful air gun, a steel ball is shot vertically upward with a velocity of 80 m/s, followed by another shot after 5 seconds. find the initial velocity of the second ball to meet the first ball 150 m from the ground.

Answer 1

first ball:

hf=vi1t-4.9t^2
second ball
hf=vi2(t-5)-4.9 (t-5)^2

you know vi1, hf. You do not know t, vi2.

in the first equation, solve for t.
then, in the second equation, solve for vi2

Answer 2

THE ANSWER IS NEGATIVE VELOCITY WHAT THE PAK

Answer 3

[AI-generated answer — Step-by-Step Bot]

To find the initial velocity of the second ball, we can use the equation of motion for the first ball:

h = ut - (1/2)gt^2

Here, h is the final height (150 m), u is the initial velocity of the first ball (80 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time (5 s).

Using this equation, we can find the time it takes for the first ball to reach a height of 150 m:

150 = 80(5) - (1/2)(9.8)(5^2)

Simplifying the equation gives:

150 = 400 - 122.5

Subtracting 400 from both sides gives:

-250 = -122.5

Dividing both sides by -122.5 gives:

t = (-250) / (-122.5) = 2.04 s

Therefore, it takes approximately 2.04 seconds for the first ball to reach a height of 150 m.

Now, let's find the initial velocity of the second ball.

Using the equation of motion for the second ball:

150 = u(2.04) - (1/2)(9.8)(2.04^2)

Simplifying the equation gives:

150 = 2.04u - 20.1

Adding 20.1 to both sides gives:

2.04u = 170.1

Dividing both sides by 2.04 gives:

u = 83.43 m/s (rounded to two decimal places)

Therefore, the initial velocity of the second ball should be approximately 83.43 m/s to meet the first ball 150 m from the ground.

Answer 4

[AI-generated answer — Explain Bot]

To find the initial velocity of the second ball, we can use the three kinematic equations of motion:

1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as

Let's assume the positive direction is upward, so the acceleration due to gravity is -9.8 m/s².

For the first ball:
Initial velocity (u1) = 80 m/s (given)
Final velocity (v1) = 0 m/s (since it reaches maximum height)
Time taken to reach maximum height (t1) = ?
Displacement (s1) = ?

Using Equation 1, we have v1 = u1 + at1, where a = -9.8 m/s² and v1 = 0 m/s.
0 = 80 - 9.8t1
9.8t1 = 80
t1 = 80 / 9.8
t1 ≈ 8.16 s

Now, using Equation 2, we can find the displacement s1:
s1 = u1t1 + (1/2)at1^2
s1 = 80 * 8.16 + (1/2) * (-9.8) * (8.16)^2
s1 ≈ 329.68 m

Since the maximum height reached by the first ball is not 150 m or higher, the second ball will collide with the first ball before reaching maximum height.

For the second ball:
Initial velocity (u2) = ?
Final velocity (v2) = -80 m/s (since it is moving downward)
Time taken to meet the first ball (t2) = t1 - 5 s
Displacement (s2) = 150 m

Using Equation 2, we have s2 = u2(t1 - 5) + (1/2)(-9.8)(t1 - 5)^2
150 = u2(8.16 - 5) + (1/2)(-9.8)(8.16 - 5)^2
150 = u2(3.16) - 4.9(3.16)^2
150 = 3.16u2 - 4.9(9.9856)
150 = 3.16u2 - 48.825184
3.16u2 = 150 + 48.825184
3.16u2 = 198.825184
u2 = 198.825184 / 3.16
u2 ≈ 63.04 m/s

Therefore, the initial velocity of the second ball should be approximately 63.04 m/s in order to meet the first ball 150 m from the ground.