A basketball player throws the ball at a 38° angle above the horizontal to a hoop which is located a horizontal distance L = 5.6 m from the point of release and at a height h = 0.4 m above it. What is the required speed if the basketball is to reach the hoop?
To find the required speed of the basketball, we can use the principles of projectile motion. Projectile motion is the motion of an object moving in a curved path under the influence of gravity.
The first step is to break down the initial velocity of the basketball into horizontal and vertical components. The vertical component governs the height reached by the basketball, while the horizontal component governs the distance traveled.
In this case, the basketball is thrown at an angle of 38° above the horizontal. The horizontal component of the initial velocity (Vx) remains constant throughout the motion and is given by:
Vx = V * cosθ
where V is the magnitude of the initial velocity and θ is the angle of projection.
The vertical component of the initial velocity (Vy) changes continuously due to the influence of gravity. To calculate the vertical component, we can use the following formula:
Vy = V * sinθ
Now, let's analyze the vertical motion of the basketball. We know that the initial vertical velocity is Vy, the final vertical displacement is h (0.4 m), and the acceleration due to gravity is -9.8 m/s^2 (taking downwards as negative).
Using the kinematic equation:
Δy = (Vy * t) + (0.5 * a * t^2)
where Δy is the vertical displacement, Vy is the initial vertical velocity, t is the time, and a is the acceleration.
By substituting the given values, we get:
0.4 = (V * sin38°) * t + (0.5 * -9.8 * t^2)
Simplifying the equation, we have:
0.4 = Vt * 0.615 + -4.9t^2
Now, let's analyze the horizontal motion of the basketball. We know that the horizontal displacement is L (5.6 m) and the horizontal velocity is Vx.
Using the equation:
Δx = Vx * t
and substituting for Vx, we have:
5.6 = V * cos38° * t
Now, we have two equations and two unknowns (V and t). By solving this system of equations simultaneously, we can find the required speed (V) if the basketball is to reach the hoop.
First, solve the second equation for t:
t = 5.6 / (V * cos38°)
Substitute this expression for t in the first equation:
0.4 = V * sin38° * (5.6 / (V * cos38°)) + -4.9 * (5.6 / (V * cos38°))^2
Simplify the equation:
0.4 = 5.6 * tan38° + -4.9 * (5.6 / V)^2
Rearrange the equation:
(5.6 / V)^2 = (0.4 - 5.6 * tan38°) / -4.9
Take the square root of both sides:
5.6 / V = sqrt((0.4 - 5.6 * tan38°) / -4.9)
Now, solve for V:
V = 5.6 / sqrt((0.4 - 5.6 * tan38°) / -4.9)
Using a calculator, plug in the value of tan38° and evaluate the expression to find the required speed for the basketball to reach the hoop.
Range = Vo^2*sin2A/g = 5.6 m.
Vo^2*sin76/9.8 = 5.6
.099Vo^2 = 5.6
Vo^2 = 56.56
Vo = 7.52 m/s.