A basketball player 2.08 m tall wants to make
a basket from a distance of 16 m. The hoop is
at a height of 3.05 m.
The acceleration of gravity is 9.8 m/s2 .
Neglect air friction.
2.08 m
3.05 m
16 m
v0
40.7◦
If he shoots the ball (from a height of
2.08 m) at a 40.7 ◦ angle, at what initial
speed must he throw the basketball so that
it goes through the hoop without striking the
backboard?
Answer in units of m/s.
To find the initial speed at which the basketball player should throw the ball, we can use the principles of projectile motion.
First, let's break down the given information:
- The basketball player's height (h) is 2.08 m.
- The height of the hoop (H) is 3.05 m.
- The distance from the player to the hoop (d) is 16 m.
- The angle at which the player throws the ball (θ) is 40.7°.
In projectile motion, the vertical and horizontal components are independent.
1. Vertical Component:
The vertical motion is affected by gravity. We can use the following equation to calculate the time it takes for the ball to reach the hoop height:
H = h + v0sin(θ)t - (1/2)gt^2
where v0 is the initial speed, and g is the acceleration due to gravity.
Since the ball reaches the same height as the hoop, the equation becomes:
H = (1/2)gt^2
Solving for t:
t = sqrt((2H)/g)
2. Horizontal Component:
The horizontal motion is not affected by gravity. We can use the following equation to calculate the horizontal distance covered by the ball:
d = v0cos(θ)t
Solving for v0:
v0 = d / (cos(θ)t)
Now, we have the values of H, h, g, θ, and d. We can substitute these values into the equations to find v0.
Calculating:
- H = 3.05 m
- h = 2.08 m
- g = 9.8 m/s^2
- θ = 40.7°
- d = 16 m
1. Calculate t:
t = sqrt((2H)/g)
t = sqrt((2 * 3.05) / 9.8)
t ≈ 0.4926 s (rounded to 4 decimal places)
2. Calculate v0:
v0 = d / (cos(θ)t)
v0 = 16 / (cos(40.7°) * 0.4926)
v0 ≈ 23.18 m/s (rounded to 2 decimal places)
Therefore, the basketball player needs to throw the ball at an initial speed of approximately 23.18 m/s for it to go through the hoop without striking the backboard.