Three balls are launched simultaneously, as in the figure. The red ball is launc

Question

Three balls are launched simultaneously, as in the figure. The red ball is launched at an initial velocity v vector_0 directed at an angle theta above the horizontal. The blue ball is launched vertically with a speed of 11.2 m/s, and the yellow ball is launched horizontally with a speed of 4.30 m/s. The blue and red balls reach identical maximum heights and return to the ground at the same time, and the red ball returns to the ground to meet the yellow ball at the same place along the horizontal. Find (a) the launch speed v_0 and (b) the launch angle theta for the red ball. (a) Number Units (b) Number Units

Explanation / Answer

This cannot be resolved unless we know how far above the ground the yellow ball is launched. If it is launched from ground level, then the red ball must have been fired vertically as well as all three balls will strike the ground at the same spot.

If this is just an oversight and the yellow ball is launched a height H above the ground, then the problem looks like this.

This assumes that the ground is level
For the red and blue balls to reach the same height and to land at the same time, they must have the same initial vertical velocity.

The time needed for that vertical velocity to be reduced to zero at the maximum height is
(0 - vy) = gt
t = vy/g
an equal amount of time will be needed to return the red and blue balls to the ground. The total flight time is
t = 2vy/g
Meanwhile, the yellow ball takes how much time to hit the ground?
y = ½gt²
t = (2y/g)

so where the vertical drop is the distance H that the yellow ball originates.
t = (2H/g)

And travels how far horizontally in that time (ignoring air resistance)
d = vt
d = 4.3*(2H/g)

The horizontal velocity of the red ball must create the same distance in the red ball's flight time

d = vx(t)
4.3*(2H/g) = vx(2vy/g)
vx = 4.3(2H/g) / (2vy/g)
vx = 2.15(2Hg) / vy

so the firing angle for the red ball is   tan = vy/vx
tan = vy / 2.15(2Hg) / vy
tan = vy² / 2.15(2Hg)
tan = 11.2² / 2.15(2H*9.81)
tan = 13.17/H

and the launch speed for the red ball is
s = (vx² + vy²)
s = ((2.15(2Hg) / vy)² + vy²)
s = ((2.15(2H(9.81) / 11.2)² + 11.2²)
s = (0.723*H + 125.44)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.