A basketball player is standing on the floor 10.0 m from the basket as in the figure below. The height of the basket is H = 3.05 m, and he shoots the ball at an angle, θ = 37.6°, with the horizontal from a height of h = 2.20 m.



(a) What is the acceleration of the basketball at the highest point in its trajectory? [__]magnitude [downwards] direction

(b)At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?

To solve this problem, we will need to analyze the motion of the basketball projectile. We can break down the problem into two parts: finding the acceleration at the highest point and finding the initial speed required to make the ball pass through the hoop.

(a) To find the acceleration at the highest point of the projectile's trajectory, we can consider the vertical motion of the basketball. At the highest point, the vertical velocity component will be zero, which means the ball is momentarily at rest before falling back down. We can use the kinematic equation for vertical motion to find the acceleration at this point.

The kinematic equation for vertical displacement is given by:
Δy = v0y * t + (1/2) * a * t^2

where Δy is the vertical displacement, v0y is the initial vertical velocity, t is the time, and a is the acceleration.

At the highest point, Δy will be equal to the vertical displacement from the starting point to the apex of the trajectory, which is given by h. So, Δy = h = 2.20 m.

Since the ball is momentarily at rest at the highest point, v0y = 0.

Applying these values, the kinematic equation simplifies to:
h = (1/2) * a * t^2

We can solve this equation for the acceleration, a:
a = (2 * h) / t^2

The time it takes for the ball to reach the highest point is the same as the time it takes for the ball to fall back down to the floor. This time can be calculated using the equation for time of flight of a projectile:

t = (2 * v0 * sin(θ)) / g

where v0 is the initial speed, θ is the launch angle, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting this value of t into the equation for acceleration, we get:
a = (2 * h) / [(2 * v0 * sin(θ) / g)^2]

Simplifying this equation further, we have:
a = (g^2 * h) / (v0^2 * sin^2(θ))

Now, we substitute the given values: h = 2.20 m, g = 9.8 m/s^2, and θ = 37.6°. This will give us the magnitude of the acceleration.

To determine the direction, we can analyze the vertical motion. Since the ball is at its highest point, it is moving upwards before it starts to fall back down due to gravity. Therefore, the acceleration of the basketball at the highest point is directed downward.

(b) To determine the initial speed required to make the ball pass through the hoop without striking the backboard, we need to analyze the horizontal motion of the basketball.

The horizontal displacement of the ball from its initial position to the basket is given as 10.0 m.

The horizontal distance traveled by the ball can be calculated using the equation for horizontal displacement:

x = v0x * t

where x is the horizontal displacement, v0x is the initial horizontal velocity, and t is the time of flight (which we can calculate as mentioned above).

Since there is no horizontal acceleration (assuming negligible air resistance), the initial horizontal velocity v0x remains constant throughout the projectile motion.

From this equation, we can solve for v0x:
v0x = x / t

Substituting the given values of x = 10.0 m and t from the previous calculations, we can determine the minimum initial speed required for the ball to pass through the hoop without hitting the backboard.

To solve this problem, we will use the equations of motion and kinematic equations.

(a) First, let's determine the time it takes for the basketball to reach the highest point in its trajectory. We can use the equation:

h = (1/2) * g * t^2

Where h is the initial height (2.20 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time. Rearranging the equation, we get:

t = sqrt((2 * h) / g)
= sqrt((2 * 2.20) / 9.8)
= sqrt(0.4489)
= 0.6699 seconds (rounded to four decimal places)

Now, we can find the initial vertical velocity of the basketball using the equation:

v_y = g * t

v_y = 9.8 * 0.6699
= 6.565 ms (rounded to three decimal places)

At the highest point of its trajectory, the vertical velocity becomes zero while the acceleration due to gravity remains constant. Therefore, the acceleration of the basketball at the highest point is -9.8 m/s^2 (downwards) since gravity acts in the downward direction.

(a) The acceleration of the basketball at the highest point in its trajectory is 9.8 m/s^2 in the downward direction.

(b) To find the speed at which the player must throw the basketball, we can use the equations of motion in the horizontal direction. The range of the projectile can be determined using the equation:

R = (v^2 * sin(2θ)) / g

Where R is the range, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. We know the range is equal to 10 meters, θ is 37.6°, and g is 9.8 m/s^2. Rearranging the equation, we get:

v = sqrt((R * g) / sin(2θ))
= sqrt((10 * 9.8) / sin(74.6))
= sqrt(98 / 0.9603)
= sqrt(101.89)
= 10.094 ms (rounded to three decimal places)

Therefore, the player must throw the basketball with a speed of 10.094 m/s in order for the ball to go through the hoop without striking the backboard.

nadh