Question

1. An object that is in flight after being thrown or projected is called a projectile. such a projectile might be a football a cricket ball, a baseball, or any other object. The motion of a projectile may be thought of as the result of two-component. One component is along the horizontal direction with no acceleration and another vertical direction with constant acceleration. i. What is the equation of the path of a projectile?
ii. What Is the time of maximum height?
iii. Formula for the time of maximum height.
iv. What is the horizontal range v. Formula for horizontal range.​

Answer 1

Answer:

An object that is in flight after being thrown or projected is called a projectile. such a projectile might be a football a cricket ball, a baseball, or any other object. The motion of a projectile may be thought of as the result of two-component. One component is along the horizontal direction with no acceleration and another vertical direction with constant acceleration. i. What is the equation of the path of a projectile?

ii. What Is the time of maximum height?

iii. Formula for the time of maximum height.

iv. What is the horizontal range v. Formula for horizontal range.

Answer 2

For answering each part, let us assume the following:

Let a projectile of mass m be projected from the ground making an angle Ф with the horizontal with initial velocity u. Let it take time T for finishing its trajectory. Let maximum height reached by projectile be H.

i.  What is the equation of the path of a projectile?

• Let us use the formula s= ut + $\frac{1}{2}$at²

          ⇒   y = ut + $\frac{1}{2}$ (-g) t²

Now initial velocity u has 2 components :

1. ucosФ in x-axis
2. usinФ in y-axis

Also horizontal distance covered be

                ⇒ x = ucosФ ×T   (distance = speed × time)

                ⇒ T = $\frac{x}{ucosФ}$

⇒ y = usinФ × $\frac{x}{ucosФ}$ - $\frac{1}{2}$g ($\frac{x}{ucosФ}$)²

⇒ y = xtanФ− $\frac{gx^{2} }{2u^{2}cos^{2} }$ is the equation of path of projectile.

ii. What Is the time of maximum height?

• At maximum height H, vertical component of velocity is 0.
• Applying v = u +at

                   ⇒  0 = usinФ -gt

                   ⇒ usinФ =gt

                  ⇒ t = $\frac{ usinФ}{g}$ is the time taken to reach maximum height H.

iii.Formula for the time of maximum height.

• At maximum height H, vertical component of velocity is 0.

   Applying v² - u² =2as

                  ⇒ 0² - (usinФ)² =2(-g)H

                  ⇒ u² sinФ =2gH

                  ⇒ H = $\frac{u²sin²Ф}{2g}$ is equation of maximum height.

iv. What is the horizontal range?

• Horizontal Range (R) is the maximum displacement of the projectile in its horizontal direction.
• Physically it is the distance between point of projection and point of return to the ground.

v. Formula for horizontal range.​

• As discussed before, horizontal distance covered is

          R = ucosФ × T

  ⇒ R = ucosФ × $\frac{2usin}{g}$        (Total time = 2 x time taken to reach max height)

  ⇒ R = $\frac{2u^{2}sin. cos }{g}$

  ⇒ R =$\frac{u^{2}sin2theta }{g}$                 (sin2Ф = 2sinФcosФ)

Thus formula for horizontal range R is $\frac{u^{2} sin2 }{g}$.