The ball and the hoop accelerate downward equally so aim the ball right at the hoop :)
ball initial vertical speed Vi = 22 sin T
ball v = Vi - g t
ball height = Vi t - (1/2)g t^2
ball u = 22 cos T
31 m = u t
hoop height = 52 - (1/2)g t^2
when they meet
Vi t -(1/2)g t^2 = 52 -(1/2)g t^2
so
Vi t = 52
or
22 sin T * t = 52
meanwhile in the horizontal direction
22 cos T * t = 31
sin T = 52/(22t)
cos T = 31/(22t)
tan T = 52/31
T = 59.2 degrees
t = 31 /22 cos T = 2.75 seconds in air
remember at the crash
hoop height = 52 - (1/2)g t^2
so
52 - 4.9 (2.75)^2
= 52 - 37.1
= 14.9 sure enough
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 22 m/s. At the exact moment the thrower releases the apple, the other performer drops the hoop. The hoop falls straight down. (Assume
d = 31 m and h = 52 m.
Neglect the height at which the apple is thrown.)At what height above the ground does the apple go through the hoop? the answer key is 14.9m but I don't know how to get there nor which equation to use for this problem. d is the distance on the ground from the performer throwing the apple on the left to the other performer, who is holding hoop on the right and h is the height from the ground up to the hoop.
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