Question

A player kicks a football with an initial speed of 25 ms\(^{-1}\) at an angle of 45\(^{\circ}\) from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion? (Take g = 10 ms\(^{-2}\))

• A. h\(_{\text{max}}\) = 10 m T = 2.5 s
• B. h\(_{\text{max}}\) = 15.625 m T = 3.54 s
• C. h\(_{\text{max}}\) = 15.625 m T = 1.77 s
• D. h\(_{\text{max}}\) = 3.54 m T = 0.125 s

Hint:

In projectile motion, always resolve the initial velocity into horizontal (\(u\cos\theta\)) and vertical (\(u\sin\theta\)) components. The vertical motion determines the time of flight and maximum height, while the horizontal motion is uniform (constant velocity, assuming no air resistance). The time to reach the maximum height is when the vertical velocity becomes zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the initial velocity and angle of projection of a football. We need to find the maximum height it reaches and the time it takes to get there. This is a standard projectile motion problem.
Step 2: Key Formula or Approach:
For a projectile launched with initial speed \(u\) at an angle \(\theta\) with the horizontal:
1. Time to reach maximum height: \( T = \frac{u \sin\theta}{g} \)
2. Maximum height: \( h_{\text{max}} = \frac{(u \sin\theta)^2}{2g} \)
Step 3: Detailed Explanation:
Given values:
Initial speed, \( u = 25 \, \text{m/s} \)
Angle of projection, \( \theta = 45^{\circ} \)
Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
First, let's calculate the time taken to reach the highest point (T).
\[ T = \frac{u \sin\theta}{g} \] \[ T = \frac{25 \sin(45^{\circ})}{10} \] We know that \( \sin(45^{\circ}) = \frac{1}{\sqrt{2}} \approx 0.707 \).
\[ T = \frac{25 \times \frac{1}{\sqrt{2}}}{10} = \frac{2.5}{\sqrt{2}} \] \[ T \approx \frac{2.5}{1.414} \approx 1.767 \, \text{s} \] Rounding to two decimal places, \( T \approx 1.77 \, \text{s} \).
Next, let's calculate the maximum height (\(h_{\text{max}}\)).
\[ h_{\text{max}} = \frac{(u \sin\theta)^2}{2g} \] We can also write this as \( h_{\text{max}} = \frac{u_y^2}{2g} \), where \( u_y = u\sin\theta \) is the initial vertical velocity.
\( u_y = 25 \sin(45^{\circ}) = 25 \times \frac{1}{\sqrt{2}} \)
\[ h_{\text{max}} = \frac{\left(25 \times \frac{1}{\sqrt{2}}\right)^2}{2 \times 10} \] \[ h_{\text{max}} = \frac{25^2 \times (\frac{1}{\sqrt{2}})^2}{20} \] \[ h_{\text{max}} = \frac{625 \times \frac{1}{2}}{20} = \frac{625}{40} \] To simplify the fraction, divide numerator and denominator by 5: \( \frac{125}{8} \).
\[ h_{\text{max}} = 15.625 \, \text{m} \] Step 4: Final Answer:
The maximum height reached is 15.625 m, and the time taken to reach it is 1.77 s. This corresponds to option (C).