Assume: A 78 g basketball is launched at an angle of 46.7
◦
and a distance of 18.6 m from
the basketball goal. The ball is released at the
same height (ten feet) as the basketball goal’s
height.
A basketball player tries to make a long
jump-shot as described above.
The acceleration of gravity is 9.8 m/s
2
.
What speed must the player give the ball?
Answer in units of m/s
L=vₒ²•sin2α/g,
vₒ=sqrt(Lg/ sin2α) =sqrt(18.6•93.8/sin93.4) =13.51 m/s
To find the speed that the player must give the ball, we can use the equations of projectile motion.
First, let's break down the given information:
- Mass of the basketball (m) = 78 g = 0.078 kg
- Launch angle (θ) = 46.7 degrees
- Distance from the goal (d) = 18.6 m
- The ball is released at the same height as the goal's height (which we can assume to be 10 feet or approximately 3.048 m)
- Acceleration due to gravity (g) = 9.8 m/s^2
Now, let's find the speed (v) of the basketball. We can use the following equation:
d = (v^2 * sin(2θ)) / g
Rearranging the equation, we can solve for v:
v^2 = (d * g) / sin(2θ)
v = √((d * g) / sin(2θ))
Plugging in the values:
v = √((18.6 * 9.8) / sin(2(46.7)))
Simplifying further:
v ≈ √(181.428 / 0.832)
v ≈ √218.567
v ≈ 14.785 m/s
Therefore, the speed that the player must give the ball is approximately 14.785 m/s.
To determine the speed at which the player must give the ball, we can use the principles of projectile motion. Here are the steps to solve for the speed:
Step 1: Convert the angle from degrees to radians.
To convert from degrees to radians, we use the formula:
angle in radians = (angle in degrees) × (π / 180)
In this case, the angle in radians would be:
angle in radians = 46.7° × (π / 180) = 0.816 radians (rounded to 3 decimal places)
Step 2: Use the vertical motion equations to find the time of flight.
The vertical motion of the basketball can be described by the equation:
y = v₀y * t + (1/2) * g * t²
Since the basketball starts and lands at the same height, the initial and final vertical displacements will be zero, and the equation becomes:
(1/2) * g * t² = 0
Simplifying the equation, we find:
t² = 0
t = 0
Step 3: Use the horizontal motion equation to find the time of flight.
The horizontal motion of the basketball can be described by the equation:
x = v₀x * t
In this case, the horizontal displacement is given as 18.6 m, and we need to solve for the time (t).
t = x / v₀x
Step 4: Use the relationship between the horizontal and vertical velocities to solve for the initial velocity (v₀).
The relationship between the horizontal and vertical velocities can be described by the equation:
v₀y = v₀ * sin(θ)
v₀x = v₀ * cos(θ)
In this case, v₀y = 0 (since the basketball is not initially moving vertically), so we can use the equation:
tan(θ) = v₀y / v₀x
Substituting the values we have, we get:
tan(0.816) = 0 / v₀x
Simplifying the equation, we find:
v₀x = 0
Since v₀x = v₀ * cos(θ), this implies that the initial velocity in the x-direction is zero.
Step 5: Use the vertical motion equation to solve for the initial vertical velocity (v₀y).
The vertical motion of the basketball can be described by the equation:
y = v₀y * t + (1/2) * g * t²
Substituting the values we have, we get:
0 = v₀y * 0 + (1/2) * 9.8 * 0²
Simplifying the equation, we find:
0 = 0
Since the equation simplifies to zero, this implies that the initial vertical velocity in the y-direction is zero.
Step 6: Use the time of flight and the vertical motion equation to solve for the initial velocity (v₀).
Since the time of flight is zero (t = 0), the equation simplifies to:
0 = v₀y * 0 + (1/2) * 9.8 * 0²
Since this equation also simplifies to zero, we cannot determine the initial velocity (v₀)
Based on the given information, it is not possible to determine the speed at which the player must give the ball.