Animated Solution for Physics - Kinematics: A player kicks a football with an initial speed of $25{\text{ ms}}^{-1}$ 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, $g=10{\text{ ms}}^{-2}$)

Visualizing the Kick

Imagine you are standing on a football field. You kick the ball with an initial speed of $u=25\text{ m/s}$ at an angle of $\theta =45^{\circ }$. The ball flies through the air, tracing a beautiful parabolic path before hitting the ground.

We need to find two crucial pieces of information: the maximum height it reaches, and the exact time it takes to get to that highest point. Let's break this down step by step.

Let's tackle the maximum height first. At the very peak of its flight, the football momentarily stops moving upwards. Its vertical velocity becomes exactly zero. We can directly use the standard formula for the maximum height of a projectile, which is derived from the third equation of motion in the vertical direction:

Now, let's carefully substitute our known values into this formula. The initial speed $u$ is $25$, the angle of projection $\theta$ is $45^{\circ }$, and the acceleration due to gravity $g$ is given as $10{\text{ m/s}}^2$.

Let's simplify this step by step. $25^2$ gives us $625$. The sine of $45^{\circ }$ is $\frac{1}{\sqrt{2}}$. When we square that, the square root disappears, leaving us with exactly $\frac{1}{2}$. And the denominator is simply $2\times 10$, which is $20$.

If we do the division, we get exactly $15.625\text{ m}$. That is the maximum height the football reaches!

The Free-Fall Shortcut

Half the problem is solved! Next, we need the time taken to reach this highest point. Here is a brilliant shortcut. Due to the symmetry of projectile motion, the time taken to go up to the maximum height is exactly the same as the time it would take to free-fall from that height.

So, we can simply use the free-fall equation:

Rearranging for time $t$, we get:

Let's use the maximum height we just found. We substitute $15.625$ for $H$, and $10$ for $g$. This avoids recalculating the sine components and makes our life much easier.

Multiplying $15.625$ by $2$ gives us exactly $31.25$. Dividing that by $10$ shifts the decimal point, leaving us with the square root of $3.125$.

Rounding it off to two decimal places, we get $1.77\text{ s}$. Looking at our options, option (c) perfectly matches both our calculated values. This was a classic application of projectile motion formulas. Always remember to carefully separate the horizontal and vertical components of motion!