A basketball player shoots the ball from a height of 2.22 m above the floor at a horizontal distance of 1 m from the basket which is 3.05 m above the floor. If she shoots the ball at an angle of 42° above the horizontal, what initial speed must she throw the basketball so that it goes through the hoop without striking the backboard?
To find the initial speed required for the basketball to go through the hoop without striking the backboard, we can use the concepts of projectile motion.
Let's break down the given information:
- The initial height of the basketball, h1 = 2.22 m
- The height of the basket, h2 = 3.05 m
- The horizontal distance from the shooter to the basket, range = 1 m
- The angle above the horizontal, θ = 42°
To solve for the initial speed, we can use the following equation of motion that relates the distance traveled horizontally, the launch angle, and the initial speed:
range = (initial speed^2 * sin(2θ)) / g
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging the equation, we can solve for the initial speed:
initial speed^2 = (range * g) / sin(2θ)
Finally, taking the square root of both sides will give us the initial speed:
initial speed = sqrt((range * g) / sin(2θ))
Plugging in the values:
range = 1 m
g = 9.8 m/s^2
θ = 42°
initial speed = sqrt((1 * 9.8) / sin(2 * 42))
Calculating sin(2 * 42) ≈ 0.9076:
initial speed = sqrt((1 * 9.8) / 0.9076)
Simplifying:
initial speed = sqrt(10.98 / 0.9076)
Calculating the square root:
initial speed ≈ sqrt(12.09)
Therefore, the initial speed at which the basketball must be thrown is approximately 3.48 m/s.