Question

A circus performer throws an apple toward a hoop held by a performer on a platform (see figure below). The thrower aims for the hoop and throws with a speed of 28 m/s. At the exact moment the thrower releases the apple, the other performer drops the hoop. The hoop falls straight down. (Assume
d = 27 m and h = 54 m.
Neglect the height at which the apple is thrown.)
(a) At what height above the ground does the apple go through the hoop?

m

Answer 1

I assume that d is the horizontal distance from the thrower to the spot below the hoop.

so, throwing it directly toward the hoop's initial position, the angle θ is such that

tanθ = 54/27 = 2

So, the horizontal speed is
vx = 28 cosθ = 28/√5 = 12.52 m/s
So, it takes the apple 27/12.52 = 2.156 seconds to reach the falling hoop.

Now, the hoop at time 2.156 has fallen to a height of

54 - 4.9*2.156^2 = 31.223 m

extra credit: do the constraints work out? What is the apple's height at t=2.156?

Answer 2

Steve can you explain what is the square root of 5 at the 28/square root of 5 is?