A basketball player is standing on the floor 10.0 m from the basket as in the fi

Question

A basketball player is standing on the floor 10.0 m from the basket as in the figure below. The height of the basket is H = 3.05 m,and he shoots the ball at an angle, = 42.8°, with the horizontal from a height of h = 1.98 m.

(a) What is the acceleration of the basketball at the highest 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 backboard?
m/s

Explanation / Answer

here,

a)

the acceleration of the basketball at the highest point in its trajectory is g = 9.81 m/s^2 and acting DOWNWARDS

b)

horizontal distance , x = 10 m

vertical distance , y = H - h = 1.07 m

theta = 42.8 degree

let the initial speed of the ball be v

using equation of trajectory

y = x * tan(theta) - g * x^2 /( 2 * v^2 * cos^2(theta))

1.07 = 10 * tan(42.8) - 9.81 * 10^2 /( 2 * v^2 * cos^2(42.8))

solving for v

v = 10.5 m/s

the initial speed of the ball is 10.5 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.