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

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

$y=x\tan \theta -\frac{1}{2}\frac{gx^2}{u^2\cos ^2\theta }$

substituting the values we get

$4.55=40\tan (60)-0.5\times \frac{9.8\times (40)^2}{u^2\cdot \cos ^260 }$

$4.55=69.28-0.5\times \frac{15,680}{u^2\cdot 0.25}$

$\frac{31,360}{u^2}=69.28-4.55$

$\frac{31,360}{64.73}=u^2$

$u^2=484.47$

$u=22.01\ m/s$

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