Answer to Question #3181 in Mechanics | Relativity for zizi

Question

A basketball player is standing on the floor 10.0 m from the basket .The height of the basket is 3.05 m and he shoots the ball at 40.0 angle with the horizantal from the height of 2.00 m .
A- What is the acceleration of the of the basketball at the heights 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 bachboard?

Answer 1

First, establish a rectangualr coordinate system, with the origin at say 2m above the floor and 10m from the hoop.

Initial position: (x_i, y_i) = (0,2)

Final position: (x_f, y_f) = (10, 1.05) ...note that y_f is 1.05 m above the origin defined above.

Launch angle is 40 deg above horizontal.

To find the initial speed, we find the components of the initial velocity, whcih we write

V_ix = V_i cos 40 V_iy = V_i sin 40

horizontal Velocity = V*Cosθ
If T is the total time of flight, and the basket is 10m away then you know that V*Cosθ*T = 10m

time required to reach the maximum height
Initial velocity/g = V*Sinθ/g = T₁

Now T₁ is only the part of the problem, because Total time = Time to rise (T₁) plus time to fall down(T₂).

Max height :(V*Sinθ)² = 2gH
(Remember H is 2m higher than the ground.)
How long to drop from max height?
(H - 1.02) = 1/2 g* T₂²
(The 1.02 is the difference in height above the release point.)
1.05=v²sin40*(10/v²cos40)-0.5*9.8*(10/v²cos40)²
1.05=10tan40 - (490/v²(cos40)²)
v²((cos40)²*10tan40) - 490 = v²((cos40)²)
v²((cos40)²*10tan40) - v²((cos40)²) = 490
v²(4.337215) = 490
v = (490/4.337215)^0.5
v = 10.66m/s
Now acceleration a = -g, where g = 9.81 m/s² is the gravity acceleration which is directed verticaly to the ground, horizonlal component of acceleration=0.