A basketball player is standing on the floor 10.0 m from
the basket as in Figure P4.50. The height of the basket is
3.05 m, and he shoots the ball at a 40.0o angle with the
horizontal from a height of 2.00 m. (a) What is the acceleration
of the basketball at the highest point in its trajectory?
(b) At what speed must the player throw the basketball so
that the ball goes through the hoop without striking the
backboard?
b. Dx = Vo^2*sin(2A)/g = 10 m
Vo^2*sin80/9.8 = 10
0.1Vo^2 = 10
Vo^2 = 100
Vo = 10 m/s[40o].
To find the answer to these questions, we need to use the laws of projectile motion. Projectile motion can be analyzed by splitting the motion into horizontal and vertical components.
(a) What is the acceleration of the basketball at the highest point in its trajectory?
At the highest point in its trajectory, the vertical component of the basketball's velocity becomes zero. We can use the equation for vertical velocity of a projectile to find the time it takes for the ball to reach its highest point:
v_y = v_0_y + a_y * t
Since v_y = 0 at the highest point, we have:
0 = v_0_y + (-g) * t
Solving for t gives us:
t = v_0_y / g
Now, we can find the acceleration at the highest point by using the equation for projectile motion:
y = y_0 + v_0_y * t - (1/2) * g * t^2
Since the vertical displacement at the highest point is 2.00 m, we have:
2.00 = 0 + v_0_y * (v_0_y / g) - (1/2) * g * (v_0_y / g)^2
Simplifying this equation will give us the value of the acceleration at the highest point.
(b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?
To answer this, we need to find the initial speed of the basketball. We can use the equations for horizontal and vertical displacement of a projectile:
x = x_0 + v_0_x * t
y = y_0 + v_0_y * t - (1/2) * g * t^2
Considering the horizontal motion, the basketball travels a distance of 10.0 m, so:
10.0 = 0 + v_0_x * t
Solving for v_0_x in terms of t, we get:
v_0_x = 10.0 / t
Now, considering the vertical motion, the basketball reaches a height of 3.05 m. We can use the equation for vertical displacement to find the time it takes for the ball to reach this height:
3.05 = 0 + v_0_y * t - (1/2) * g * t^2
Simplifying this equation will give us an expression for v_0_y in terms of t.
Finally, to avoid hitting the backboard, the vertical displacement at the horizontal distance of 10.0 m should be 3.05 m. We can plug in these values to find the initial speed of the basketball.
By following these steps and performing the calculations, you can find the answers to both parts (a) and (b) of the question.
To solve this problem, we can break it down into several steps:
Step 1: Find the initial velocity of the basketball.
Given:
- Angle of the shot (θ) = 40.0°
- Initial height (h) = 2.00 m
We can find the initial vertical velocity using the formula:
v₀y = v * sin(θ)
Where v₀y is the initial vertical velocity and v is the initial velocity.
Since the ball is shot horizontally (no initial vertical velocity), we have:
v₀y = 0
Thus, the initial velocity of the basketball is given by:
v = v₀x
Step 2: Find the time it takes for the basketball to reach its highest point.
To find the time it takes for the ball to reach its highest point (the maximum height), we can use the formula:
h = v₀y * t + 0.5 * g * t²
Where:
- h is the maximum height (3.05 m)
- v₀y is the initial vertical velocity at t = 0 (0 m/s)
- g is the acceleration due to gravity (-9.8 m/s²)
- t is the time
Rearranging the equation and substituting the given values, we get:
0.5 * g * t² = h
-4.9 * t² = 3.05
t² = 3.05 / -4.9
t ≈ √(-0.622)
Since time cannot be negative in this context, we can disregard the negative value. Thus, the time it takes for the basketball to reach its highest point is:
t ≈ 0.79 seconds (rounded to two decimal places)
Step 3: Find the acceleration at the highest point.
At the highest point of the projectile's trajectory, the vertical velocity is zero. Thus, we can find the vertical acceleration (a) using the formula:
v = v₀y + a * t
Where:
- v is the final vertical velocity (0 m/s)
- v₀y is the initial vertical velocity (0 m/s)
- a is the vertical acceleration (what we need to find)
- t is the time at the highest point (0.79 seconds)
Rearranging the equation, we get:
a * t = -v₀y
a ≈ -v₀y / t
≈ 0 / 0.79
≈ 0 m/s²
Therefore, the acceleration at the highest point is approximately 0 m/s².
Step 4: Find the minimum speed needed.
To determine the minimum speed at which the player must throw the basketball for it to pass through the hoop without striking the backboard, we need to consider the horizontal and vertical distances traveled by the ball.
The horizontal distance traveled by the basketball can be found using:
d = v₀x * t
Where:
- d is the horizontal distance (10.0 m)
- v₀x is the initial horizontal velocity (what we need to find)
- t is the time at the highest point (0.79 seconds)
Substituting the given values, we get:
10.0 = v₀x * 0.79
v₀x ≈ 12.66 m/s (rounded to two decimal places)
Therefore, the player must throw the basketball with a minimum speed of approximately 12.66 m/s to pass through the hoop without striking the backboard.