Juggling Class
Juggling Class
A circus performer juggling while standing on a platform 15.0 m high tosses a ball directly up…?
A circus performer juggling while standing on a platform 15.0 m high tosses a ball directly upward into the air at a speed of 5.0 m/s. If it leaves his hand 0.7 m above the platform, what is the ball’s maximum altitude? If the Juggler misses the ball, at what speed will it hit the floor? Ignore air friction.
Soooo…vi=5 vf=0 a=-9.8 m/s^2 because in our class down is negative. I’ve done it over and over and keep getting the wrong answers.
For the second question…vi=5 i think a=-9.8 m/s^2 again and delta x=(-15 + .7) all times -1 is what I think but again keep getting the wrong answer
thank you so much that is correct!! I was not using 15.7 as the delta x which is why I kept getting it wrong. My silly mistakes:) you are brilliant!
You must utilize the kinematic eqn:
y = yo + (vo)t – ½ gt²,
where yo = 15.7, and vo = 5 m/s
and also v = vo – gt, but v = 0 at max height.
vo = gt, and t = vo/g = 5/9.81 = 0.51 s
so that max height:
y = 15.7 + 5(0.51) – ½ (9.81)(0.51²)
= 18.25 – 1.274 = 16.976 m
If The Juggler misses the catch, then we use the kinematic
eqn. v² = (vo)²+2gy, where vo = 0
Therefore, the final velocity just before impact with ground is
v = √2gy = √2(9.8)(16.976) = 18.24 m/s
My Juggling Class
