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?
First, establish a rectangualr coordinate system, with the origin at say 2m above the floor and 10m from the hoop.
Initial position: (xi, yi) = (0,2)
Final position: (xf, yf) = (10, 1.05) ...note that yf 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
Vix = Vi cos 40 Viy = Vi 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.