Problema Solution
"A rock is thrown upward with a velocity of 28 meters per second from the top of a 31 meter high cliff, and it misses the cliff on the way down. When will the rock be 10 meters from the rock, below? Round your answer to two decimal places."
The formula given for this word problem is h=-1/2gt^2-(initial velocity)t+(initial height)
Answer provided by our tutors
h=-1/2gt^2-(initial velocity)t+(initial height)
if we plug the values given in the text we get:
h(t) = (-1/2)gt^2 - 28t + 31
we need to find t, t>0 such that h(t) = 31 - 10 (t0 meters bellow the cliff) that is
(-1/2)gt^2 - 28t + 31 = 31 - 10
g = 9.81 m/s^2
(-1/2)*9.81*t^2 - 28t + 31 = 31 - 10
by solving we find:
t = 0.34 s
click here to see the step by step solution of the equation:
in 0.34 seconds the rock will be 10 meters bellow the cliff.