Determining Maximum Horizontal Distance of a Projectile

Question:

The ceiling of a long hall is 25 m high.What is the maximum horizontal distance that a ball thrown with a speed of 40 ms^-1 can go without hitting the ceiling of the hall?

Solution: Calculating the maximum horizontal distance

Projectile motion image:

Maximum height of the projectile:

Maximum height of a projectile is given by

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

Here, H=25 m (Ceiling of the long hall)

Initial speed of the ball (u) = 40 ms^-1

Acceleration due to gravity is 9.8 ms^-2

$${25=\frac{(40^2)\sin^2\theta}{19.6}}$$

$${490=(1600)\sin^2\theta}$$

$${\sin^2\theta= \frac{490}{1600}}$$

$${\sin^2\theta=0.306}$$

$${\sin\theta=\sqrt(0.306)}$$

To determine the angle:

$${\sin\theta=0.553}$$

$${\theta =\sin^{-1}(0.553)}$$

$${\theta = 33.6^\circ}$$

Maximum horizontal distance of the projectile:

The horizontal distance(Range) is given by $${R=\frac{u^2\sin2\theta}{g}}$$

$${R=\frac{40^2\sin 67.2^\circ}{9.8}}$$

$${R=\frac{(1600)(0.921)}{9.8}}$$

$${R=150.36 \ m}$$

The maximum distance the ball can be thrown without hitting the ceiling is approximately 150.5 m

Projectile motion with examples:

Conclusion:

In many disciplines, including physics, engineering, sports, ballistics, aerospace, mathematics, and computer science, the projectile motion and its equations are crucial. One precisely forecasts 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, like 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.