A basketball player throws the ball at a 47degree angle above the horizontal to a hoop which is located a horizontal distance L = 8.4 m from the point of release and at a height h = 2.7 m above it. What is the required speed if the basketball is to reach the hoop?
I'm trying to do it, but I'm not sure how to.
If we assume that the ball just grazes the hoop and somehow falls in, we can work it out like that.
In real life, of course, the ball will rise above the hoop and fall in from somewhere above it, but since no such information was given, we have to just assume that
y(0) = 0
y(8.4) = 2.7
Our equation for the trajectory is
y = x tanθ - g/(2 (v cosθ)^2) x^2
Pluggin our numbers, we have
y = x tan 47° - 4.9/(v cos47°)^2 x^2
= 1.072x - 10.535/v^2 x^2
So, since y(8.4) = 2.7, we have
1.072(8.4) - 10.535/v^2 (8.4^2) = 2.7
v = 10.858 m/s
To find the required speed for the basketball to reach the hoop, we can use the equations of motion in projectile motion. Here are the steps to solve the problem:
Step 1: Split the initial velocity into horizontal and vertical components. The initial velocity (V0) can be separated into its horizontal component (V0x) and vertical component (V0y). The horizontal component remains constant throughout the motion, while the vertical component changes due to the effect of gravity.
Step 2: Use the horizontal component to determine the time of flight. The time of flight (t) is the duration the basketball is in the air. Since only the horizontal component affects the horizontal motion, the time of flight can be calculated using the horizontal distance and horizontal component of the velocity (V0x).
Step 3: Use the vertical component to determine the maximum height. The maximum height (H) reached by the basketball can be found using the vertical component of velocity (V0y). The time taken to reach the maximum height is half the total time of flight (t/2).
Step 4: Use the vertical component to determine the total time of flight. The total time of flight (t) can be found using the vertical component of velocity (V0y) and the initial height (h) using the equation: h = V0y * t - (1/2) * g * t^2, where g is the acceleration due to gravity.
Step 5: Use the total time of flight to find the vertical component of velocity (V0y). Since the vertical component changes with time due to gravity, we can find it using the equation: V0y = g * t, where g is the acceleration due to gravity.
Step 6: Use the desired angle and the vertical component of velocity (V0y) to calculate the initial velocity (V0). The angle (θ) and the vertical component of velocity (V0y) form a right-angled triangle. We can use the equation: sin(θ) = V0y / V0 to find the initial velocity (V0).
Step 7: Substitute the given values into the equations and solve for the required speed (V0). Given angle (θ = 47 degrees), horizontal distance (L = 8.4 m), and vertical height (h = 2.7 m), substitute these values into the equations from the previous steps and solve for the required speed (V0).
By following these steps, you should be able to find the required speed for the basketball to reach the hoop.