Question

Suppose the kicker launches the ball at 60∘ instead of 30∘. Assuming that the goal is 4.55 m high and 40 m away, what minimum initial speed v0 would the ball need to have in order to just clear the goal?

Answer 1

Answer:22 m/s

Step-by-step explanation:

Given

launch angle
`\theta =60^(\circ)`

height of goal
`h=4.55\ m`

and horizontal distance
`x=40\ m`

Suppose initial speed is
`u`

Trajectory of a Projectile is given by

`y=x\tan \theta -(1)/(2)(gx^2)/(u^2\cos ^2\theta )`

substituting the values we get

`4.55=40\tan (60)-0.5* (9.8* (40)^2)/(u^2\cdot \cos ^260 )`

`4.55=69.28-0.5* (15,680)/(u^2\cdot 0.25)`

`(31,360)/(u^2)=69.28-4.55`

`(31,360)/(64.73)=u^2`

`u^2=484.47`

`u=22.01\ m/s`

So, initial launch speed is
`22\ m/s`