If we assume that the ball just grazes the hoop and somehow falls in, we can work it out like that.
In real life, of course, the ball will rise above the hoop and fall in from somewhere above it, but since no such information was given, we have to just assume that
y(0) = 0
y(8.4) = 2.7
Our equation for the trajectory is
y = x tanθ - g/(2 (v cosθ)^2) x^2
Pluggin our numbers, we have
y = x tan 47° - 4.9/(v cos47°)^2 x^2
= 1.072x - 10.535/v^2 x^2
So, since y(8.4) = 2.7, we have
1.072(8.4) - 10.535/v^2 (8.4^2) = 2.7
v = 10.858 m/s
A basketball player throws the ball at a 47degree angle above the horizontal to a hoop which is located a horizontal distance L = 8.4 m from the point of release and at a height h = 2.7 m above it. What is the required speed if the basketball is to reach the hoop?
I'm trying to do it, but I'm not sure how to.
1 answer