Part A A basketball player stands in the corner of the court at the three-point

Question

Part A A basketball player stands in the corner of the court at the three-point line, 6.37 m from the basket, with the hoop 3.30 m above the floor. If the player shoots the ball from a height of 1.92 m at 31above the horizontal, what should the launch speed be to make the basket? m/s Submit Request Answer Part B How much would the launch speed have to increase to make the ball travel 45 cm farther and miss the hoop entirely? Express your answer using two significant figures. = Submit Request Answer

Explanation / Answer

here,

the horizontal distance , x = 6.37 m

the vertical height , h = 3.3 m

a)

initial height , h0 = 1.92 m

theta= 31 degree

let the launch speed be v1

using equation of trajectory

(h- h0) = x * tan(theta) - g * x^2 /( 2 * v1^2 * cos^2(theta))

( 3.3 - 1.92) = 6.37 * tan(31) - 9.81 * 6.37^2 /(2 * v1^2 * cos^2(31))

solving for v1

v1 = 10.5 m/s

the launch speed is 10.5 m/s

b)

when x1 = x + 0.45 m = 6.82 m

let the launch speed be v2

using equation of trajectory

(h- h0) = x * tan(theta) - g * x1^2 /( 2 * v2^2 * cos^2(theta))

( 3.3 - 1.92) = 6.82 * tan(31) - 9.81 * 6.82^2 /(2 * v2^2 * cos^2(31))

solving for v2

v2 = 10.7 m/s

% = (v2 - v1)/v1 * 100

% = ( 10.7 - 10.5)/10.5 * 100

% = 1.9 %

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