Q3. Please answer all the questions and show your work clearly. Thank you. A bas

Question

Q3. Please answer all the questions and show your work clearly. Thank you.

A basketball player shoots a jump shot and scores (the ball does not touch the rim or the board). The initial velocity of the basketball is 8.20 m/s at an angle of 48 degree above the horizontal. The release point of the basketball is 2.40 m above the floor. The basket is 3.05 m above the floor. (a) What are the horizontal (x) and vertical (y) components of the initial velocity? (b) What is the horizontal distance between the player and the basket? (c) Find the final velocity of the basketball when it enters the hoop, including both the magnitude and direction (with respect to the horizontal).

Explanation / Answer

(a)

horizonal component v0x = v0*costheta = 8.2*cos48 = 5.48 m/s

vertical component voy = v0*sintheta = 8.2*sin48 = 6.09 m/s

(b)

PROJECTILE

along horizontal

initial velocity vox = vo*costheta

ax = 0

from equation of motion

x-Xo = vox*T+ 0.5*ax*T^2

x-X0 = vo*costheta*T

T = (x-X0)/(vo*costheta)......(1)

along vertical

voy = vo*sintheta

acceleration ay = -Eq/m = 6.4*10^5*1.6*10^-19/(1.67*10^-27) = -6.13*10^13 m/s^2

from equation of motion

y-y0 = voy*T + 0.5*ay*T^2

y-y0 = (vo*sintheta_0*(x-x0))/(vo*costheta_0) + (0.5*ay*(x-x0)^2)/(vo^2*(costheta_0)^2)

y-y0 = (x-X0)*tantheta_0 + ((0.5*ay*(x-X0)^2)/(vo^2*(costheta_0)^2))

y0 = 2.4 m

y = 3.05 m

x - x0 = ?

theta = 48

v0 = 8.2 m/s

3.05-2.4 = ((x-x0)*tan48) - (0.5*9.8*(x-x0)^2/(8.2^2*(cos48)^2)

x-x0 = 6.18 m

----------------------

(c)

time taken T = (x-x0)/vox = 6.18/5.48 = 1.13 s

vy = voy + ay*T

vy = 6.09 - (9.81*1.13)

vy = -4.99 m/s

vx = vox = 5.48 m/s

magnitude = sqrt(5.48^2+4.99^2) = 7.41 m/s

direction = tan^-1(Vy/Vx) = -42.3 degrees

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