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3 2. A block of mass m = 2.00 kg is released from rest at h = 0.500 m above the surface of a table, at the top of a = 30.0 in
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Q.2A) Using conservation of energy

\frac{1}{2}mv^2 = mgh

v= \sqrt{2gh} \\ v = \sqrt{2 \times 9.8 \ \rm m/s^2 \times 0.500 \ \rm m} \\ v = 3.13 \ \rm m/s300 below the horizontal.

Q2B) Let us first calculate the time taken by the block to reach the ground

\\ H = v\sin \theta t + \frac{1}{2}gt^2 \\ 2.0 \ \rm m = 3.13 \ \rm m/s \sin 30^{\circ} \times t+ \frac{1}{2} \times 9.8 \ \rm m/s^2 \times t^2 \\ 4.9t^2 + 1.565 t -2 = 0 \\

Solving this equation

t = 0.50 \ \rm s, -0.82 \ \rm s

Taking the positive value of time only

\\ R = v\cos \theta .t \\ R = 3.13 \ \rm m/s \times \cos30^{\circ} \times 0.50 \ \rm s \\ R = 1.36 \ \rm m

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