Assume: A 78 g basketball is launched at an

angle of 54.5

and a distance of 10.4 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.

the mass is not a factor

horizontal speed = Sh = S cos(54.5)

the flight time to the basket is
... 10.4 m / Sh = 10.4 * cos(54.5) / S

vertical speed = Sv = S sin(54.5)

gravity slows the vertical speed to zero at the peak, halfway through the flight

Sv / g = 1/2 * 10.4 / Sh

S / [g * sin(54.5)] =
... 5.2 * cos(54.5) / S

S^2 = 5.2 * g * sin(54.5) * cos(54.5)

To find the speed the player must give the basketball, we can start by analyzing the vertical motion of the ball. Since the ball is launched at the same height as the basketball goal's height, it means the initial vertical velocity of the ball is zero.

The vertical motion can be described using the equation:

y = y0 + v0yt - (1/2)gt^2

where:
- y is the vertical displacement (height difference between the starting point and the highest point of trajectory)
- y0 is the initial vertical position (which is zero because the ball is launched at the same height as the goal)
- v0y is the initial vertical velocity (which is zero)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time it takes for the ball to reach the highest point of the trajectory (which is half of the total time of flight)

Since the total time of flight is symmetrical, we can find the time it takes for the ball to reach the highest point by dividing the total time of flight by 2.

The total time of flight can be calculated using the horizontal motion of the ball. We can use the range formula to find the total time of flight.

Range (R) = v0x × t

where:
- R is the range or horizontal displacement (10.4 m)
- v0x is the initial horizontal velocity

To find the initial horizontal velocity, we can use the fact that the ball is launched at an angle of 54.5 degrees. The initial horizontal velocity (v0x) and initial vertical velocity (v0y) are related as follows:

v0x = v0cosθ
v0y = v0sinθ

where:
- v0 is the initial velocity of the ball
- θ is the launch angle (54.5 degrees)

We can rearrange the range equation to solve for the total time of flight:

t = R / v0x

Substituting in the values we have:

t = 10.4 m / (v0 * cos(54.5◦))

Now we can substitute the value of t back into the vertical motion equation:

y = (1/2)gt^2

Solving for y:

y = (1/2)(9.8 m/s^2)(10.4 m / (v0 * cos(54.5◦)))^2

Now we can calculate the vertical displacement (y) by subtracting the initial height of the ball (which is 10 feet, or approximately 3.048 meters) from the highest point of the trajectory (which is the same as the vertical displacement):

y = y - y0

Finally, we can find the initial velocity (v0) by rearranging the equation:

v0^2 = v0x^2 + v0y^2

Substituting the values we have:

v0^2 = (v0 * cos(54.5◦))^2 + (v0 * sin(54.5◦))^2

v0 = sqrt((y - y0) * 2g / (1 - cos^2(54.5◦) - sin^2(54.5◦)))

Evaluating the equation with the given values:

v0 = sqrt((3.048 m) * 2 * (9.8 m/s^2) / (1 - cos^2(54.5◦) - sin^2(54.5◦)))

Simplifying the equation:

v0 = sqrt((3.048 m) * 2 * (9.8 m/s^2) / (1 - cos^2(54.5◦) - sin^2(54.5◦)))

v0 = sqrt(5.96464 m^2/s^2)

v0 ≈ 2.441 m/s

Therefore, the player must give the ball a speed of approximately 2.441 m/s.