Question

. A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00m above the ground. At the same instant the balloon is released; a second balloon is thrown straight down at 2.00 m/s from the same height. Determine which balloon hits the ground first and how much sooner it hits the ground than the other balloon

Answer 1

Second balloon hits ground Δt = 0.185 seconds sooner than first balloon

Step-by-step explanation:

Given:-

- The first balloon is thrown horizontally with speed, u1 = 2.0 m/s

- The second balloon is thrown down with speed, u2 = 2.0 m/s

- The height from which balloon are thrown, si = 6.0 m (above ground)

Find:-

Determine which balloon hits the ground first and how much sooner it hits the ground than the other balloon

Solution:-

- We will first determine the time taken (t1) for the first balloon thrown horizontally with speed u1 = 2.0 m/s from top of building from a height of s = 6.0 m from ground to it the ground.

- Using the second kinematic equation of motion in vertical direction:

si = 0.5*g*t1^2

Where, g: The gravitational constant = 9.81 m/s^2

6.0 = 4.905*t1^2

4.905*t1^2 - 6.0 = 0

- Solve the quadratic equation:

t 1 = 1.106 s

- Similarly, the time taken (t2) for the second balloon thrown down with speed u2 = 2.0 m/s from top of building from a height of s = 6.0 m from ground to it the ground.

- Using the second kinematic equation of motion in vertical direction:

si = u2*t2 + 0.5*g*t1^2

Where, g: The gravitational constant = 9.81 m/s^2

6.0 = 4.905*t1^2 + 2*t2

4.905*t1^2 + 2*t2 - 6.0 = 0

- Solve the quadratic equation:

t 2 = 0.9208 s

- We see that the second balloon thrown down vertically hits the ground first. The second balloon reaches ground, t1 - t2 = 0.185 seconds, sooner than first balloon.

Answer 2

Step-by-step explanation:

Height of building

H = 6m

Horizontal speed of first balloon

U1x = 2m/s

Second ballot is thrown straight downward at a speed of

U2y = 2m/s

Time each gallon hits the ground

Balloon 1.

Using equation of free fall

H = Uoy•t + ½gt²

Uox = 0 since the body does not have vertical component of velocity

6 = ½ × 9.8t²

6 = 4.9t²

t² = 6 / 4.9

t² = 1.224

t = √1.224

t = 1.11 seconds

For second balloon

H = Uoy•t + ½gt²

6 = 2t + ½ × 9.8t²

6 = 2t + 4.9t²

4.9t² + 2t —6 = 0

Using formula method to solve the quadratic equation

Check attachment

From the solution we see that,

t = 0.9211 and t = -1.329

We will discard the negative value of time since time can't be negative here

So the second balloon get to the ground after t ≈ 0.92 seconds

Conclusion

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.