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: Understanding the Question:
A cricket ball is hit and caught 2 seconds later at the same elevation, representing a full projectile motion flight. We need to find its maximum height.

Step 2: Key Formula or Approach:
The total time of flight for a projectile is $T = \frac{2u\sin\theta}{g}$.
The maximum height reached is $H = \frac{u^2\sin^2\theta}{2g}$.

Step 3: Detailed Explanation:
Given the time of flight $T = 2\ \text{s}$.
Use the time of flight formula to find the vertical component of the initial velocity ($u\sin\theta$):
$$2 = \frac{2u\sin\theta}{10}$$
$$20 = 2u\sin\theta \implies u\sin\theta = 10\ \text{m/s}$$
Now, substitute this vertical velocity component into the maximum height formula:
$$H = \frac{(u\sin\theta)^2}{2g}$$
$$H = \frac{(10)^2}{2 \times 10}$$
$$H = \frac{100}{20} = 5\ \text{m}$$

Step 4: Final Answer:
The maximum height reached is $5\ \text{m}$, matching option (B).