Question

A cricket player hit a ball like a projectile but the fielder caught the ball after 2 second. The maximum height reached by a ball is ($g = 10\ \text{m/s}^2$)

• A. $2\ \text{m}$
• B. $5\ \text{m}$
• C. $4\ \text{m}$
• D. $3\ \text{m}$

Hint:

The time it takes to reach maximum height is exactly half the total time of flight ($t = 1\ \text{s}$). Using simple 1D kinematics from the peak downward ($H = \frac{1}{2}gt^2$), $H = 0.5 \times 10 \times 1^2 = 5\ \text{m}$. No trig required!

Answer 1

The Correct Option is B

Step 1: Understand the question.
A cricket ball is hit like a projectile and caught after $2$ seconds at the same height. We must find the maximum height the ball reached.

Step 2: Pick the key idea.
The whole flight takes $2$ seconds. A projectile goes up and comes back down in equal times. So it takes half the time, $1$ second, to reach the top. At the top, the upward speed is zero.

Step 3: Use the time of flight to find vertical speed.
The time of flight is $T = \dfrac{2u\sin\theta}{g}$, where $u\sin\theta$ is the upward part of the launch speed. Put $T = 2$ and $g = 10$:
\[ 2 = \frac{2\,u\sin\theta}{10} \]

Step 4: Solve for the vertical speed.
\[ 2\times 10 = 2\,u\sin\theta \quad\Rightarrow\quad u\sin\theta = 10\ \text{m/s} \]

Step 5: Use the maximum height formula.
The maximum height depends only on the upward speed:
\[ H = \frac{(u\sin\theta)^2}{2g} \]

Step 6: Put in the numbers.
\[ H = \frac{(10)^2}{2\times 10} = \frac{100}{20} = 5\ \text{m} \]
This matches option (2).
\[ \boxed{H = 5\ \text{m}} \]