A juggler traveling in a train on level track throws a ball straight up, relativ

Question

A juggler traveling in a train on level track throws a ball straight up, relative to the train, with a speed of 6.06 m/s. The train has a velocity of 19 m/s due east. As observed by the juggler, what is the ball's total time of flight? What is the displacement of the ball during its rise? According to a friend standing on the ground next to the track, what is the ball's initial speed? What is the angle of the launch? degree above the horizontal What is the displacement of the ball during its rise? x-Component: m y-component: m

Explanation / Answer

Given data:
speed = 6.06m/s
velocity = 19m/s
a)v = u + a t
t = ( v - u ) / a
where v = 6.06 m/s ; u = 0; a = -9.8m/s
t = ( 6.06- 0 ) / 9.8 0.618 s
so the total time in flight = 2*0.618 s=1.236 s
horizontal component of displacement = Vx t = 19 * 0.618 =11.742 m
vertical component of displacement = Sy = ut + 1/*a*t^2
Sy = 6.06*0.618 + 0.5 *( 9.8 )* 0.618^2=1.8736 m
total displacement = ( 11.742 + 1.8736 ) 13.6156 m
B)observed velocity = sqrt(( 19^2 + 6.06^2 )) m/s =19.94 m/s
angle theta = arc tan(6.06/19) = 17.689 degrees
c)displacement
x-component = horizontal velocity *time = 19*0.618 =11.742m/s
y-component = 6.06*0.618 = 3.745m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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