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Answer #1

here,

the mass of snowball , m = 1.1 kg

the height of cliff , h0 = 14.8 m

initial velocity , v0 = 18.5 m/s

theta = 33 degree

a)

let the speed at the bottom of the ground be v

using conservation of energy

0.5 * m * v0^2 + m * g * h0 = 0.5 * m * v^2

0.5 * 18.5^2 + 9.81 * 14.8 = 0.5 * v^2

solving for v

v = 25.2 m/s

the speed of the snowball at the bottom is 25.2 m/s

b)

when the launch angle is below the horizontal

let the speed at the bottom of the ground be v

using conservation of energy

0.5 * m * v0^2 + m * g * h0 = 0.5 * m * v^2

0.5 * 18.5^2 + 9.81 * 14.8 = 0.5 * v^2

solving for v

v = 25.2 m/s

the speed of the snowball at the bottom is 25.2 m/s


c)

the speed does not depends upon the mass of ball

so, the speed will be the same .i.e 25.2 m/s

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