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Step-by-step Solution
Step 1: Understand the Problem
A balloon is moving upward with a uniform velocity of 10 m/s. At a certain instant, when the balloon is at a height of 75 m above the ground, an object is dropped. We need to find the height of the balloon from the ground at the instant the dropped object strikes the ground. Gravity acts downward with an acceleration of 10 m/s2.
Step 2: Determine the Time Taken by the Object to Hit the Ground
Let the time taken by the object to strike the ground be $ T $. At the moment the object is dropped, it has an initial upward velocity of 10 m/s (same as the balloon). We choose upward as the positive direction.
Initial height of the object from ground, $ y_0 = 75\text{ m} $
Initial velocity of the object, $ v_0 = +10\text{ m/s} $
Acceleration due to gravity, $ a = -10\text{ m/s}^2 $
Using the equation of motion for displacement with uniform acceleration:
$ y = y_0 + v_0\,T + \tfrac{1}{2} a\,T^2 $
At the time the object hits the ground, $ y = 0 $ (taking ground as reference), so:
$ 0 = 75 + 10T - 5T^2 $
This simplifies to:
$ -5T^2 + 10T + 75 = 0 $
Dividing through by $ -5 $:
$ T^2 - 2T - 15 = 0 $
Step 3: Solve the Quadratic Equation
The solutions for $ T^2 - 2T - 15 = 0 $ are given by:
$ T = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm \sqrt{64}}{2} = \frac{2 \pm 8}{2} $
Hence, $ T = 5 \text{ s} $ (taking the positive root) or $ T = -3 \text{ s} $ (not physically meaningful, so we discard it).
Step 4: Calculate the Balloon’s Height When the Object Hits the Ground
In the same time $ T = 5\text{ s} $, the balloon continues to ascend at its constant velocity of 10 m/s. Thus, the balloon travels an additional height of:
$ \Delta h = 10 \times 5 = 50\text{ m} $
Therefore, the balloon’s height above the ground at that instant is:
$ 75\text{ m} + 50\text{ m} = 125\text{ m} $
Answer
The height of the balloon from the ground when the object strikes the ground is 125 m.
