Question

The initial speed of a cannon ball is 200 m/s. If the ball is to strike a target that is at a horizontal distance of 3.0 km from the cannon, what is the minimum time of flight for the ball

Answer 1

The minimum time of flight for the ball is 16.38 s

Step-by-step explanation:

Projectile Motion: When we throw an object which make an angle with horizontal and only one acceleration which work on that object that is acceleration due to gravity. This type of motion is known as projectile motion.

Range: The distance is covered horizontally by the object is known as Range of the object.

`R=(u^2sin2\theta)/(g)`

u = initial velocity= 200m/s

g = acceleration due to gravity = 9.8m/s²

θ= angle

R= range = 3.0 km= 3000 m

`\therefore 3000=(200^2* sin2\theta)/(9.8)`

`\Rightarrow sin 2\theta =(9.8* 3000)/(40000)`

⇒θ= 23.65°

Flight time

`T=(2usin \theta)/(9.8)`

`\Rightarrow T=\frac{2* 200 sin {23.65}}{9.8}`

`\Rightarrow T=16.38` s

The minimum time of flight for the ball is 16.38 s