Question

1 the ground #1 A ball is thrown horizontally off Cliff OF of height, h, with a velocity Vo = 5.90 mis and Strikes with a velocity, V=23.1 mis. What is the height of the cliff? * let g -9.80 mists and give your answer the correct number of sig tigs in meters.

Answer 1

the final velocity is the magnitude of x-component and y-component

v = sqrt ( v_x² + v_y² )

where v_x = v_o = 5.90 m/s

v_y = v_oy - gt

as v_oy = 0

so,

v_y = - gt

so,

23.1 = sqrt ( v_x² + v_y² )

squaring both sides, we have

533.61 = v_x² + v_y²

533.61 = 5.90² + ( -gt)²

498.8 = 96.04 * t²

t = 2.278 seconds

so,

h = 1/2 * g * t²

h = 25.44 meters

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