An Olympic basketball player shoots towards a basket that is 5.66 m horizontally from her and 3.05 m above the floor. The ball leaves her hand 1.74 m above the floor at an angle of 42.0o above the horizontal. What initial speed should she give the ball so that it reaches the basket and hopefully scores?
I found the answer to be 10.75
I know that I am 4 yrs late but could u still please see if my answer is right or not...
plzzz
To find the initial speed required for the ball to reach the basket, we can use the equations of projectile motion.
First, let's break down the given information:
- Horizontal distance (range) = 5.66 m
- Vertical distance (height) = 3.05 m
- Initial vertical position = 1.74 m
- Angle of projection (θ) = 42.0°
To solve this problem, we'll use the following equations for projectile motion:
1. Range equation:
Range = (initial speed^2 * sin(2θ))/g
2. Maximum height equation:
Maximum height = (initial speed^2 * sin²(θ))/(2g)
3. Vertical displacement equation:
Vertical displacement = (initial speed^2 * sin²(θ))/(2g) - (initial vertical position)
Here, g represents the acceleration due to gravity, which is 9.8 m/s².
We can use the range equation to find the initial speed. Rearranging the equation, we have:
(initial speed)^2 = (Range * g) / sin(2θ)
Let's substitute the known values into the equation:
Range = 5.66 m
θ = 42.0°
g = 9.8 m/s²
After substituting the known values, we have:
(initial speed)^2 = (5.66 * 9.8) / sin(2 * 42.0°)
Now, we can calculate the initial speed by taking the square root of both sides:
initial speed = √[(5.66 * 9.8) / sin(2 * 42.0°)]
Calculating this value, the initial speed should be approximately 7.55 m/s.
Therefore, the Olympic basketball player needs to give the ball an initial speed of approximately 7.55 m/s for it to reach the basket and hopefully score.