Question

Three identical balls are thrown from the same height off a tall building. They are all thrown with the same speed, but in different directions. Ball A is thrown in the horizontally, parallel to the roof; ball B is thrown at an angle 45° above the horizontal; ball C is thrown at an angle 45° below the horizontal. Which ball has the greatest speed just before it hits the ground? Assume that the ground near the building is level, and ignore any effects due to air resistance.

Answer 1

Answer:

three balls have the same speed

Explanation:

In a parabolic motion we have that

$v_{x}=v_{0}cos\alpha\\v_{y}=-gt+v_{0}sin\alpha\\v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

but the time just before the balls hit the ground is

$t=\frac{2v_{0}sin\alpha}{g}$

Hence we have

ball A

$v=\sqrt{v_{0}^{2}cos^{2}(0)+(-g\frac{2v_{0}sin(0)}{g}+v_{0}sin(0))^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

ball B

$v=\sqrt{v_{0}^{2}cos^{2}(45)+(-g\frac{2v_{0}sin(45)}{g}+v_{0}sin(45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

ball C

$v=\sqrt{v_{0}^{2}cos^{2}(-45)+(-g\frac{2v_{0}sin(-45)}{g}+v_{0}sin(-45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

Hence, all three balls have the same speed just before hit the groug

vA=vB=vC

I hope this is useful for you

regards