Question

After being assaulted by flying cannonballs, the knights on the castle walls (12 m above the ground) respond by propelling flaming pitch balls at their assailants. One ball lands on the ground at a distance of 50 m from the castle walls. If it was launched at an angle of 53° above the horizontal, what was its initial speed?

Answer 1

Answer:

The answer is 24,9 mts/seg

Explanation:

The attachment shows a schematic of the situation.

Due we don't have the flying time of the projectile, we can use, From the projectile motion equations the next expression:

$y = x tan\alpha - \frac{g x^{2}}{2 Vo^{2} cos\alpha}$

Where:

y is the altitude of the wall

x is the distance from the wall where the ball landed

$\alpha$ is the shoot angle

Vo is the initial velocity

Clearing Vo, we obtain:

Vo = $\sqrt{\frac{g x^{2}}{2 cos^{2}\alpha(x tan\alpha - y)}}$

Knowing that y = 12 mts; x = 50 mts and $\alpha$ = 53° and gravity constant is 9.8 mts/seg2; the only thing that we must do is replace the values on the formula:

Vo = $Vo = \sqrt{\frac{(9.8)(50^{2})}{2cos^{2}53(50tan53-12)}}$

Vo = 24,9 mts/seg