A 78 g basketball is launched at an angle of 45.4◦ and a distance of 10.1 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/s2 . What speed must the player give the ball?
Range = Vo^2*sin(2A)/g = 10.1 m.
Vo^2*sin(90.8)/9.8 = 10.1
0.102Vo^2 = 10.1
Vo^2 = 98.99
Vo = 9.95 m/s.
To find the speed that the player must give the ball, we can use the equations of projectile motion.
Step 1: Determine the initial vertical velocity (Vy0) of the basketball.
We are told that the ball is released at the same height as the basketball goal's height, so the initial vertical velocity (Vy0) is zero.
Step 2: Determine the time it takes for the ball to reach the goal height.
The height of the basketball goal is given in feet, so we need to convert it to meters. We will use the conversion factor 1 ft = 0.3048 m.
Height of the basketball goal = 10 ft = 10 x 0.3048 m = 3.048 m
We can use the equation h = Vy0t + (1/2)gt^2, where h is the height, Vy0 is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
Since Vy0 is zero, the equation simplifies to h = (1/2)gt^2.
Plugging in the values, we have 3.048 = (1/2)(9.8)t^2.
Simplifying the equation:
3.048 = 4.9t^2
t^2 = 3.048 / 4.9
t^2 = 0.62245
t = √0.62245
t ≈ 0.7888 s
Step 3: Determine the horizontal velocity (Vx0) of the basketball.
The horizontal distance traveled by the basketball (10.1 m) is equal to Vx0 * t, where Vx0 is the initial horizontal velocity and t is the time.
Rearranging the equation, we have Vx0 = (distance) / t.
Plugging in the values, we have Vx0 = 10.1 m / 0.7888 s ≈ 12.81 m/s.
Step 4: Determine the total initial velocity (V0) of the basketball.
The total initial velocity (V0) of the basketball can be found using the Pythagorean theorem.
The Pythagorean theorem states that V0 = √(Vx0^2 + Vy0^2).
Since Vy0 is zero, the equation simplifies to V0 = √(Vx0^2).
Plugging in the value, we have V0 = √(12.81^2) ≈ 12.81 m/s.
Therefore, the player must give the ball a speed of approximately 12.81 m/s.
To find the speed the player must give the ball, we can use the equations of motion for projectile motion. First, we need to determine the vertical and horizontal components of the ball's initial velocity.
Given:
Mass of the basketball (m) = 78 g = 0.078 kg
Launch angle (θ) = 45.4°
Distance to the basketball goal (d) = 10.1 m
Acceleration due to gravity (g) = 9.8 m/s^2
Step 1: Compute the vertical component of the initial velocity.
The initial vertical velocity component (Vy) can be determined using the equation:
Vy = V * sin(θ)
Where V is the initial velocity and θ is the launch angle.
Step 2: Compute the time of flight (t).
The time of flight (t) can be determined using the equation:
t = 2 * Vy / g
Step 3: Compute the horizontal component of the initial velocity (Vx).
The horizontal component of the initial velocity (Vx) can be determined using the equation:
Vx = d / t
Step 4: Compute the total initial velocity (V).
The total initial velocity (V) can be determined using the equation:
V = Vx / cos(θ)
Finally, we can calculate the speed the player must give the ball (initial velocity):
1. Calculate Vy:
Vy = V * sin(θ)
2. Calculate the time of flight (t):
t = 2 * Vy / g
3. Calculate Vx:
Vx = d / t
4. Calculate V:
V = Vx / cos(θ)
The resulting value of V will be the initial speed that the player must give the ball.