General Science Model Test Questions 27 Physics (12)

The General Science Model Test (GSMT) is a popular preparation tool for schoollevel exams. Question27 belongs to the Physics section of the 12question set. Below you will find the full statement of the question, the correct answer, and a stepbystep explanation that helps you understand the concepts behind it.

Question27 Full Statement

Answer

Explanation

To solve the problem we can use the principle of conservation of mechanical energy or kinematic equations. Both methods give the same result; the energy approach is often quicker for upanddown problems.

Method1 Energy Conservation

• Initial mechanical energy (at the point of release):
  • Kinetic energy, \(K_i = \frac12 m v_0^{2}\) where \(v_0 = 20\; \text{ms}^{-1}\).
  • Potential energy, \(U_i = m g h\) where \(h = 15\; \text{m}\) (height of the building).
• Final mechanical energy (just before hitting the ground):
  • Kinetic energy, \(K_f = \frac12 m v_f^{2}\) (what we need).
  • Potential energy, \(U_f = 0\) because the reference level is ground level.
• Set initial total energy equal to final total energy (no air resistance): \[ \frac12 m v_0^{2}+mgh = \frac12 m v_f^{2} \]
• Cancel the mass \(m\) and solve for \(v_f\): \[ v_f^{2}=v_0^{2}+2gh = 20^{2}+2(10)(15)=400+300=700 \] \[ v_f = \sqrt{700}\approx 26.5\;\text{ms}^{-1} \]
• Because the ball falls past the starting point, it gains an additional 20ms of speed (the speed it had when it returned to the launch height). Adding the two contributions: \[ v_{\text{ground}} = \sqrt{v_0^{2}+2g(15)} + v_0 = \sqrt{400+300}+20 \approx 26.5+20 \approx 46.5\;\text{ms}^{-1} \] However the above doublecounts; the correct way is to compute directly from the top of the building to the ground, which we already did. The mistake lies in mixing frames. The proper result using energy alone is: \[ v_f = \sqrt{v_0^{2}+2g(15)} = \sqrt{400+300}= \sqrt{700}\approx 26.5\;\text{ms}^{-1} \] Since the answer choices are whole numbers, the nearest listed value is **30ms**, which matches the answer key.

Method2 Kinematic Equation

• Use the equation \(v^{2}=u^{2}+2as\) where:
  • u = 20ms (upward, taken as positive)
  • a = 10ms (gravity, opposite to upward direction)
  • s = 15m (displacement from the launch point to the ground is downward)
• Substituting: \[ v^{2}=20^{2}+2(-10)(-15)=400+300=700 \] \[ v = \sqrt{700}\approx 26.5\;\text{ms}^{-1} \]
• Again the numerical result is about 26ms; the nearest option in the multiplechoice list is 30ms, so we select D.

Key Concepts Tested

• Projectile motion in one dimension: Understanding that upward motion decelerates under gravity, then reverses direction.
• Conservation of mechanical energy: Kinetic+potential energy remains constant when nonconservative forces (like air resistance) are neglected.
• Choosing the correct sign convention: Positive upward, negative downward, or viceversa, must be consistent throughout the calculation.
• Using the appropriate kinematic formula: \(v^{2}=u^{2}+2as\) is ideal when time is not required.

Common Mistakes

• Adding the speed at the top of the building to the speed gained during the fall this double counts the motion.
• Using the wrong sign for acceleration or displacement, which leads to a subtraction instead of addition.
• Forgetting to square the velocities when applying the energy or kinematic equations.
• Choosing an answer based on the exact numeric value (26.5ms) when the test expects the nearest listed option.

Study Tips for Similar Questions

• Write down what you know: initial speed, height, and the value of g.
• Decide whether the problem is easier with energy methods (no time needed) or with kinematics (specific direction matters).
• Keep a consistent sign convention and stick to it throughout the solution.
• After calculating, compare your numeric answer with the provided options; if the exact number is not listed, choose the closest reasonable value.

Mastering questions like GSMT27 builds confidence for the entire Physics section of the General Science Model Test. Practice similar upanddown problems, and you will quickly recognize the pattern: the speed just before impact is found by adding the square of the initial speed to twice the product of g and the vertical drop.