10. A man throws a ball vertically up in the air with an initial velocity V1 such that it reaches a height of 12 m with a speed of 12 m/s. If he throws the same ball vertically up with an initial velocity V2 such that it reaches a maximum height of 12 m. Calculate v₁/v₂. (up to 2 decimal places)
Calculate V₁/V₂ for a Ball Thrown Vertically Upward: Detailed Solution
Understanding the Vertical Motion Problem
This question compares two different vertical upward motions of the same ball. In the first case, the ball is thrown upward with an initial velocity V1. When it reaches a height of 12 m, it is still moving upward with a speed of 12 m/s. This means that 12 m is not the maximum height in the first case because the ball still has some kinetic energy and continues moving upward.
In the second case, the same ball is thrown upward with an initial velocity V2 and reaches a maximum height of exactly 12 m. At the maximum height, the instantaneous velocity of the ball becomes zero. This important difference between the two situations allows us to calculate V1 and V2 separately and then determine their ratio.
Equation of Motion Used in the Solution
For vertical motion under gravity, the most suitable kinematic equation is:
v2 = u2 + 2as
Here, v represents the final velocity at a particular height, u represents the initial velocity, a represents acceleration, and s represents vertical displacement. Since the ball is moving upward while gravity acts downward, the acceleration is a = −g. Therefore, the equation becomes:
v2 = u2 − 2gh
Rearranging this equation for the initial velocity gives:
u2 = v2 + 2gh
First Case: Calculating the Initial Velocity V₁
In the first situation, the ball is thrown vertically upward with an initial velocity V1. At a height of 12 m, its speed is still 12 m/s. Therefore, the given quantities are h = 12 m and v = 12 m/s.
Using the vertical motion equation:
v2 = V12 − 2gh
Substituting the given values:
(12)2 = V12 − 2g(12)
Therefore:
V12 = 144 + 24g
Second Case: Calculating the Initial Velocity V₂
In the second situation, the ball is thrown vertically upward with an initial velocity V2 and reaches a maximum height of exactly 12 m. At the highest point of vertical motion, the velocity of the ball becomes zero. Therefore, v = 0 and h = 12 m.
Applying the same equation of motion:
v2 = V22 − 2gh
At the maximum height, v = 0. Therefore:
0 = V22 − 2g(12)
Hence:
V22 = 24g
Calculating the Ratio V₁/V₂
We have obtained the following two equations:
V12 = 144 + 24g
V22 = 24g
Dividing the first equation by the second equation gives:
V12/V22 = (144 + 24g)/(24g)
Taking the square root on both sides:
V1/V2 = √[(144 + 24g)/(24g)]
Using the standard value of acceleration due to gravity, g = 9.8 m/s2:
V1/V2 = √[(144 + 24 × 9.8)/(24 × 9.8)]
V1/V2 = √[(144 + 235.2)/235.2]
V1/V2 = √(379.2/235.2)
V1/V2 = √1.6122
V1/V2 ≈ 1.2697
Therefore, up to two decimal places:
V1/V2 = 1.27
Why Is V₁ Greater Than V₂?
The value of V1 must be greater than V2 because the ball in the first case still has a speed of 12 m/s after already reaching a height of 12 m. It therefore possesses both gravitational potential energy and kinetic energy at that height.
In contrast, the ball in the second case just reaches the height of 12 m and momentarily comes to rest there. Its speed at 12 m is zero, so its entire initial kinetic energy has been converted into gravitational potential energy. Consequently, the first throw requires a greater initial velocity than the second throw, which is consistent with the calculated ratio V1/V2 being greater than 1.
Alternative Solution Using Conservation of Mechanical Energy
The same result can be obtained using the law of conservation of mechanical energy. In the first case, the initial kinetic energy of the ball is converted into gravitational potential energy as well as the kinetic energy that remains when the ball reaches the height of 12 m.
Therefore:
½mV12 = mgh + ½m(12)2
Multiplying throughout by 2/m gives:
V12 = 2gh + 144
Since h = 12 m:
V12 = 24g + 144
In the second case, the ball reaches its maximum height of 12 m with zero speed. Therefore, its entire initial kinetic energy is converted into gravitational potential energy:
½mV22 = mgh
Hence:
V22 = 2gh = 24g
Dividing the two expressions and taking the square root again gives:
V1/V2 ≈ 1.27
This confirms the result obtained using the equations of motion.
Physical Interpretation of the Answer
The ratio V1/V2 = 1.27 means that the initial speed required in the first case is approximately 1.27 times the initial speed required in the second case. The additional initial speed is necessary because the ball must not only reach the height of 12 m but must also retain a speed of 12 m/s at that height.
The problem demonstrates an important principle of vertical motion: reaching a certain height with a non-zero speed requires more initial kinetic energy than merely reaching the same height as the maximum point of motion.
Final Answer
V1/V2 = 1.27
The ball thrown with initial velocity V1 still has a speed of 12 m/s at a height of 12 m, whereas the ball thrown with initial velocity V2 has zero speed at the same height because 12 m is its maximum height. Applying the equation v2 = u2 − 2gh to both cases and calculating the ratio gives V1/V2 = 1.27 up to two decimal places.
