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