here,
the initial speed , u = 21.5 m/s
theta = 15 degree
initial height , h0 = 2.33 m
height of net , h = 1.07 m
a)
let the horizontal distance be x
using equation of trajectory
(h - h0) = x * tan(theta) - g * x^2 /( 2 * u^2 * cos^2(theta))
( 1.07 - 2.33) = x * tan(15) - 9.81 * x^2 /( 2 * 21.5^2 * cos^2(15))
solving for x
x = 27.58 m
the horizontal distance is 27.58 m
b)
let the final vertical speed be vy
vy^2 - (u * sin(theta))^2 = - 2 * (h - h0) * g
vy^2 - ( 21.5 * sin(15))^2 = 2 * ( 2.33 - 1.07) * 9.81
solving for vy
vy = 7.46 m/s
vx = u * cos(theta) = 20.77 m/s
the final speed , v = sqrt(vx^2 + vy^2) = 22.1 m/s
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