Suppose a ball is thrown upward at a velocity of 44ft/sec from a cliff 200 feet above a dry riverbed. Predict the height of the ball after 3 seconds, and then determine the time that the ball will hit the riverbed.

Question

anonymous · Answer

y = (1/2)*g*t^2 +v0*t + y0

so your y0 = 200 feet.  for the first part of the question, the time = 3 and the initial velocity = 44 ft/s

gravity = -32 ft/s^2

so you have y =  (1/2)*(-32)*3^2 +44*3 + 200

therefore y = 188 (check my math with a calculator, I did it in my head.)

anonymous · Answer

$$y-y_0 =v_0t - {1 \over 2} g t^2$$