Question:

A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

Solution:Calculating the maximum height of the ball

Projectile motion:

Maximum horizontal distance(Range):

Maximum range is attained when the ball is projected at an angle of 45^o with the horizontal

$${R=\frac{u^2}{g}}$$

$${100=\frac{u^2}{g}}$$

Maximum height of the ball:

Maximum height attained by the ball from the ground is

$${H=\frac{u^2}{2g}}$$

$${H=\frac{u^2}{g}(\frac{1}{2})}$$

$${H=\frac{100}{2}}$$

$${H=50 \ m}$$

Conclusion

In many disciplines, including physics, engineering, sports, ballistics, aerospace, mathematics, and computer science, the projectile motion and its equations are crucial.One may precisely forecast a projectile’s trajectory and future movements by knowing the beginning velocity, projection angle, motion along the x and y axes, and acceleration due to gravity.

These equations are used not just for object prediction and targeting, but also for developing and studying a variety of machinery and structures, such as bridges and rocket missiles.The study of projectile motion and related equations will continue to play a crucial part in our lives with the continuous development of technology and mathematics.