What is its final velocity just before it hits the ground?
How long does it take to hit the ground?
Henry2
answered
V = ?
b. h = 0.5g*t^2 = 65.
4.9*t^2 = 65.
Step-by-Step Bot
answered
Step 1: Calculate the acceleration due to gravity.
The acceleration due to gravity is a constant value of approximately 9.8 m/s².
Step 2: Solve for the final velocity.
Using the equation of motion: v² = u² + 2as,
where v is the final velocity, u is the initial velocity (0 m/s as it was dropped), a is the acceleration due to gravity, and s is the displacement (65.0m).
v² = 0² + 2(9.8)(65.0)
v² = 0 + 1274
v ≈ √1274
v ≈ 35.7 m/s
Step 3: Solve for the time taken to hit the ground.
Using the equation of motion: s = ut + 0.5at²,
where s is the displacement (65.0m), u is the initial velocity (0 m/s), a is the acceleration due to gravity, and t is the time taken.
65.0 = 0( t) + 0.5(9.8)(t²)
65.0 = 4.9t²
t² = 65.0/ 4.9
t² ≈ 13.3
t ≈ √13.3
t ≈ 3.65 s
Therefore, the final velocity just before the water balloon hits the ground is approximately 35.7 m/s, and it takes approximately 3.65 seconds to hit the ground.
Explain Bot
answered
v = √(2gh)
Where:
v = final velocity
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height from which the object is dropped
Now, let's calculate the final velocity of the water balloon:
v = √(2 × 9.8 m/s² × 65.0 m)
v ≈ √(1274 m²/s²)
v ≈ 35.7 m/s
Therefore, the final velocity of the water balloon just before hitting the ground is approximately 35.7 m/s.
To find the time it takes for the water balloon to hit the ground, we can use the equation:
h = (1/2)gt²
Where:
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height from which the object is dropped
t = time taken to hit the ground
Rearranging the equation, we get:
t = √(2h/g)
Now, let's calculate the time it takes for the water balloon to hit the ground:
t = √(2 × 65.0 m / 9.8 m/s²)
t ≈ √(13.3 s²)
t ≈ 3.65 s
Therefore, it takes approximately 3.65 seconds for the water balloon to hit the ground.
