# Answer to Question #3181 in Mechanics | Relativity for zizi

Question #3181

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?

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?

Expert's answer

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

Initial position: (x

Final position: (x

Launch angle is 40 deg above horizontal.

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

V

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 = T

Now T

Max height :(V*Sinθ)

(Remember H is 2m higher than the ground.)

How long to drop from max height?

(H - 1.02) = 1/2 g* T

(The 1.02 is the difference in height above the release point.)

1.05=v

1.05=10tan40 - (490/v

v

v

v

v = (490/4.337215)

v = 10.66m/s

Now acceleration a = -g, where g = 9.81 m/s

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 40horizontal 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 = T

_{1}Now T

_{1}is only the part of the problem, because Total time = Time to rise (T_{1}) plus time to fall down(T_{2}).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* T

_{2}^{2}(The 1.02 is the difference in height above the release point.)

1.05=v

^{2}sin40*(10/v^{2}cos40)-0.5*9.8*(10/v^{2}cos40)^{2}1.05=10tan40 - (490/v

^{2}(cos40)^{2})v

^{2}((cos40)^{2}*10tan40) - 490 = v^{2}((cos40)^{2})v

^{2}((cos40)^{2}*10tan40) - v^{2}((cos40)^{2}) = 490v

^{2}(4.337215) = 490v = (490/4.337215)

^{0.5}v = 10.66m/s

Now acceleration a = -g, where g = 9.81 m/s

^{2}is the gravity acceleration which is directed verticaly to the ground, horizonlal component of acceleration=0.
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