a person in a hot air balloon ascending vertically at a constant rate of 12m/s drops a rock over the side when the balloon is 40m above the ground, what is the velocity of the rock just before it hits the ground, how long will it take the rock to reach the ground
a. V^2 = Vo^2 + 2g*h = (-12)^2 + 19.6*40 = 928, V = 30.5 m/s.
b. Vo + g*t = 30.5. -12 + 9.8t = 30.5, t = 4.34 s. To reach gnd.
To find the velocity of the rock just before it hits the ground, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
First, let's calculate the initial velocity of the rock. The person in the hot air balloon is ascending vertically at a constant rate of 12m/s. Since the rock is dropped from the balloon, its initial velocity will be the same as the balloon's velocity, but in the opposite direction. So, the initial velocity is -12m/s.
Next, we need to calculate the displacement of the rock. The balloon is 40m above the ground when the rock is dropped. Therefore, the displacement is 40m.
The acceleration, in this case, is due to gravity and is approximately 9.8m/s^2 (assuming no air resistance).
Now, we can substitute the values into the equation and calculate the final velocity:
v^2 = (-12)^2 + 2 * 9.8 * 40
v^2 = 144 + 2 * 9.8 * 40
v^2 = 144 + 784
v^2 = 928
v ≈ √928
v ≈ 30.46m/s (rounded to two decimal places)
So, the velocity of the rock just before it hits the ground is approximately 30.46 m/s.
To calculate the time taken for the rock to reach the ground, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = displacement (distance to be covered, 40m)
u = initial velocity (-12m/s)
t = time taken
a = acceleration due to gravity (-9.8m/s^2)
We can rearrange the equation to solve for time, t:
40 = -12t + (1/2)(-9.8)(t^2)
40 = -12t - 4.9t^2
To find the time taken, we can solve the above quadratic equation for t.
-4.9t^2 - 12t + 40 = 0
We can solve this quadratic equation using the quadratic formula or by factoring. However, in this case, the equation doesn't factor easily, so let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
t = (12 ± √((-12)^2 - 4 * (-4.9) * 40)) / (2 * (-4.9))
t = (12 ± √(144 + 784)) / (-9.8)
t = (12 ± √928) / (-9.8)
Solving this equation, we get two solutions:
t ≈ 4.45s or t ≈ -1.99s
Since time cannot be negative in this case, the time taken for the rock to reach the ground is approximately 4.45 seconds.
So, the velocity of the rock just before it hits the ground is approximately 30.46 m/s, and it takes approximately 4.45 seconds to reach the ground.