Question

the maximum horizontal range projectly is 980m find the velocity of is projection.g= 9.8m/s​

Answer 1

Answer:

Let Velocity is V and the angle of projection with horizontal is θ.

∴Range, R=

g

2V

2

sinθcosθ

​

=

g

V

2

sin2θ

​

 

⇒

dθ

dR

​

=

g

V

2

​

2cos2θ=0  ⇒2θ=90

o

    ⇒θ=45

o

⇒

dθ

2

d

2

R

​

=−

g

V

2

​

4sin2θ, at θ=45

o

  ⇒

dθ

2

d

2

R

​

=−

g

4V

2

​

<0

Hence, at θ=45

o

 , Range is maximum.