A balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant it is 64 feet above the ground.
a. How many seconds after its release will the bag strike the ground?
b. At what velocity will it hit the ground?
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To solve this problem, we will use the equation of motion for the object falling freely under gravity:
h = ut + (1/2)gt^2
where:
h is the height of the object,
u is the initial vertical velocity of the object,
t is the time,
g is the acceleration due to gravity (-32 ft/s^2, if we consider upwards as positive).
a. How many seconds after its release will the bag strike the ground?
In this case, the height of the bag when released is 64 feet above the ground, and the initial velocity of the balloon is given as 16 ft/s. Since the sandbag is initially at rest relative to the balloon, its initial upward velocity is also 16 ft/s.
Using the equation h = ut + (1/2)gt^2, we can substitute the values:
64 = 16t + (1/2)(-32)t^2
Rearranging the equation and simplifying:
32t^2 + 16t - 64 = 0
Dividing the equation by 16 to simplify further:
2t^2 + t - 4 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = 2, b = 1, and c = -4. Substituting the values:
t = (-(1) ± √((1)^2 - 4(2)(-4))) / 2(2)
Simplifying further:
t = (-1 ± √(1 + 32)) / 4
t = (-1 ± √33) / 4
Since time cannot be negative in this context, we can ignore the negative solution:
t = (-1 + √33) / 4
Therefore, the sandbag will strike the ground approximately t = 0.793 seconds after being released.
b. At what velocity will it hit the ground?
To find the velocity at which the sandbag hits the ground, we can use the formula:
v = u + gt
Since we are calculating the final velocity when it hits the ground, the height at this point is 0 and the initial upward velocity of the sandbag is 16 ft/s.
Using the equation v = u + gt, we can substitute the values:
v = 16 + (-32)(0.793)
v ≈ -7.376 ft/s (rounded to three decimal places)
Therefore, the sandbag will hit the ground with an approximate velocity of -7.376 ft/s. Note that the negative sign indicates the downward direction.
This question does not require calculus or integration.
It does require that you know the acceleration of gravity is 32 ft/s^2.
a. Solve for time t:
height = 64 + 16 t - (32/2)t^2 = 0
t^2 -t -4 = 0
t = (1/2)[1+ sqrt17] = 2.562 seconds
(ignore then negative root, -1.562 s)
b. V2 at impact = 16 - 32*2.562
= -66.0 ft/s
Energy Check: V2^2/2 - V1^2/2 = g*16 ft
as required