Answer:
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.
Answer:
Second balloon hits ground Δt = 0.185 seconds sooner than first balloon
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.
