Question

A ball is thrown upwards with a velocity of \(20\,m/s\). Find the maximum height reached \((g=10\,m/s^2)\).

• A. \(10\,m\)
• B. \(20\,m\)
• C. \(30\,m\)
• D. \(40\,m\)

Hint:

At the highest point of vertical motion, the ball is momentarily at rest. Think about how the initial kinetic energy at launch relates to the gravitational potential energy gained by the time it stops rising, and set the two equal to connect speed, gravity and height.

Answer 1

The Correct Option is B

Concept: When a body is thrown vertically upward, its velocity gradually decreases due to the downward acceleration caused by gravity. At the maximum height, the velocity becomes zero. We use the equation of motion: \[ v^2 = u^2 - 2gh \] where
• \(u\) = initial velocity
• \(v\) = final velocity
• \(g\) = acceleration due to gravity
• \(h\) = maximum height

Step 1: Write the given values. \[ u = 20\,m/s \] \[ v = 0 \quad (\text{at maximum height}) \] \[ g = 10\,m/s^2 \]

Step 2: Substitute into the equation of motion. \[ 0^2 = 20^2 - 2(10)h \] \[ 0 = 400 - 20h \]

Step 3: Solve for \(h\). \[ 20h = 400 \] \[ h = 20 \] Thus, the maximum height reached is \[ \boxed{20\,m} \]

Answer 2

Concept:

• The work-energy theorem: the work done by gravity while the ball rises equals the change in its kinetic energy.
• At the maximum height, the ball is momentarily at rest, so all of its initial kinetic energy has been converted into gravitational potential energy.
• This lets the height be found directly by equating energies, without writing out the equations of motion at all.

Step 1: Write the initial kinetic energy at launch.
$KE_i = \tfrac{1}{2}mu^2$, with $u = 20\,m/s$.

Step 2: Apply the work-energy theorem up to the highest point.
At maximum height, $v = 0$, so $KE_f = 0$. The lost kinetic energy equals the gained gravitational potential energy:
$\tfrac{1}{2}mu^2 = mgh$   (mass cancels from both sides)

Step 3: Solve for $h$.
$h = \dfrac{u^2}{2g} = \dfrac{(20)^2}{2 \times 10} = \dfrac{400}{20} = 20$

Final Answer: Maximum height $= 20\,m$.