Asked by Anonymous
If a ball is thrown off a 123 m cliff at a speed of 18 m/s, how far from the base of the diff is the ball?
Givens:
Unknown:
Equation:
Substitute:
Solve:
Givens:
Unknown:
Equation:
Substitute:
Solve:
Answers
Answered by
Henry
Xo = 18 m/s. = Hor. velocity.
h = o.5g*t^2.
123 = 4.9t^2, t = 5.01 s. to reach gnd.
d = Xo*t = 18 * 5.01 = 90.2 m.
h = o.5g*t^2.
123 = 4.9t^2, t = 5.01 s. to reach gnd.
d = Xo*t = 18 * 5.01 = 90.2 m.
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