Question

15) Derive the relation for equation of trajectory,time of flight,horizontal range and maximum height
attain when an object is thrown with some initial velocity u making an angle e​

Answer 1

Answer:

Let a projectile move with a velocity u which is inclined with the horizontal at angle of θ. The velocity after time 't' will be

v=v  

x

​  

 

i

^

+v  

y

​  

 

j

^

​  

 

v=ucosθ  

i

^

+(usinθ−gt)  

j

^

​  

 

(Since:v  

y

​  

=usinθ−gt)

​  

 

At maximum height the projectile will only have horizontal component that is  

v  

x

​  

=ucosθ

v  

y

​  

 

2

−u  

y

​  

 

2

=2ay

v  

y

​  

=0(atmaxheightH)

u  

y

​  

=usinθ

ay=−gH

Puttingthesevalues,

0=(usinθ)  

2

−2gH

H=  

2g

u  

2

sin  

2

θ

​  

 

​  

 

 

​