Question

A cannon sends a projectile towards a target a distance 1420 m away. The initial velocity makes an angle 35◦ with the horizontal. The target is hit. The acceleration of gravity is 9.8 m/s2 . What is the magnitude of the initial veloc- ity?

Answer 1

121.7 m/s

Step-by-step explanation:

Horizontal range, R = 1420 m

Angle of projection, θ = 35°

acceleration due to gravity, g = 9.8 m/s^2

let u be the velocity of projection.

Use the formula for horizontal range

`R=(u^(2)Sin2\theta )/(g)`

`1420=(u^(2)Sin70 )/(9.8)`

u = 121.7 m/s

Thus, the velocity of projection is 121.7 m/s.

Answer 2

`v_(o)=141.51m/s`

Step-by-step explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

`x=1420m\\g=-9.8m/s^(2)`

From the equation of x-position we know that

`x=v_(ox)t=v_(o)cos(35)t`

Solving for
`v_(o)`

`v_(o)=(x)/(tcos(35)) =(1420m)/(tcos(35))` (1)

Now, if we analyze the equation of y-position we got

`y=y_(o)+v_(oy)t+(1)/(2)gt^(2)`

At the end of the motion y=0

`0=v_(o)sin(35)t+(1)/(2)gt^(2)`

Knowing the equation for
`v_(o)` in (1)

`0=(1420)/(tcos(35))tsin(35)-(1)/(2)(9.8)t^(2)`

`(1)/(2)(9.8)t^(2)=1420tan(35)`

Solving for t

`t=\sqrt{(2(1420tan(35)))/(9.8) } =14.25s`

Now, we can solve (1)

`v_(o)=(1420m)/((14.25s)cos(35))=141.51m/s`