A basketball player 1.9 m tall wants to
make a basket from a distance of 7.8 m. The
hoop is at a height of 3.05 m.If he shoots the ball (from a height of 1.9 m)
at a 23.5� angle, at what initial speed must he
shoot the basketball so that it goes through
the hoop without striking the backboard?
The acceleration due to gravity is 9.8 m/s2 .
Neglect air friction.
Answer in units of m/s
To find the initial speed at which the basketball player must shoot the ball, we can use the equations of projectile motion. First, let's break down the given information:
Height of the basketball player, h = 1.9 m
Height of the hoop, H = 3.05 m
Distance to the hoop, d = 7.8 m
Angle of the shot, θ = 23.5 degrees
Acceleration due to gravity, g = 9.8 m/s^2
To solve this, we'll use the following equations of motion for projectile motion:
1. Vertical motion equation: H = h + v₀ * sin(θ) * t - 0.5 * g * t^2
2. Horizontal motion equation: d = v₀ * cos(θ) * t
Since we want to find the initial speed (v₀), and the time of flight (t) is the same in both equations, we can solve for t using the horizontal motion equation and substitute it into the vertical motion equation.
First, let's find the time of flight (t) using the horizontal motion equation:
d = v₀ * cos(θ) * t
Rearranging the equation to solve for t:
t = d / (v₀ * cos(θ))
Now, substitute this expression for t into the vertical motion equation:
H = h + (v₀ * sin(θ) * (d / (v₀ * cos(θ)))) - (0.5 * g * (d / (v₀ * cos(θ))))^2
Simplifying the equation:
H = h + (d * tan(θ)) - (0.5 * g * (d^2 / (v₀^2 * cos^2(θ))))
Rearranging the equation to isolate v₀:
0.5 * g * (d^2 / (v₀^2 * cos^2(θ))) = h + (d * tan(θ)) - H
To isolate v₀, we'll move the terms around:
v₀^2 * cos^2(θ) = (g * d^2) / (2 * (h + (d * tan(θ)) - H))
Finally, take the square root of both sides:
v₀ = sqrt((g * d^2) / (2 * (h + (d * tan(θ)) - H)))
Now, we can plug in the given values:
v₀ = sqrt((9.8 * 7.8^2) / (2 * (1.9 + (7.8 * tan(23.5)) - 3.05)))
Calculating this, we get:
v₀ ≈ 8.6 m/s
Therefore, the basketball player must shoot the basketball at an initial speed of approximately 8.6 m/s in order to make the basket without hitting the backboard.