Question

) A projectile is projected at an angle θ with the horizontal. Find an expression for

(i) time of flight (ii) maximum height reached (iii) horizontal range.​

Answer 1

Given :

A projectile is projected at an angle θ with the horizontal.

To find :

(i) time of flight

(ii) maximum height reached

(iii) horizontal range.

Solution :

⚫In 'x' direction / towards x axis :-

Distance (horizontal)= R

Velocity = U cosθ

Acceleration = 0

where:- U = initial velocity

⚫ In 'y' direction / towards y axis :-

Distance (Sy)= 0

Velocity = U Sinθ

Acceleration = -g

1. Now to find Time of flight use formula :-

⟹$s = ut + \frac{1}{2} a {t}^{2}$

put values in the Formula according to y axis :-

⟹$0 = ut \: sinθ - \frac{1}{2} g {t}^{2}$

⟹$- ut \: sinθ= - \frac{1}{2} g {t}^{2}$

⟹$u\: sinθ= \frac{1}{2} g t$

⟹$2u\: sinθ= g t$

⟹$g t =2u\: sinθ$

⟹$t = \frac{2u\: sinθ}{g }$

where, t = time period

2. Now to find maximum height , use formula:-

v²-u²= 2as

put values in the Formula according to y axis :-

v = 0

u = u sinθ

a = -g

s = H ( maximum height)

⟹0-(u sinθ)²= - 2g H

⟹u² sin²θ = 2g H

⟹(u² sin²θ)/2g = H

3. Now to find horizontal range , use formula:-

R ( horizontal range / distance) = U cosθ T

where U = initial velocity

⟹T = time of flight = (2u sinθ )/ g

⟹R =(u cosθ 2u sinθ )/ g

⟹R =(u² 2cosθ sinθ )/ g

⟹R =(u² sin2θ )/ g

ANSWER :

⟹$t = \frac{2u\: sinθ}{g }$

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

⟹$R \: = \frac{ {u}^{2} sin2θ}{g}$

where :-

t = time of flight

H = maximum height

R = horizontal range

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(Refer attachment for better understanding)