Problema Solution
A rock is thrown upward with a velocity of 27 meters per second from the top of a 35 meter sliff and it misses the cliff on the way down. When will it be 5 meters from the water?
Answer provided by our tutors
we use the following equation for location of a thrown object:
x = vt - gt^2, where 'x' is the location in meters, 'v' is velocity in m/s, 't' is time in seconds and 'g' is the force of gravity 9.8m/s^2
initial velocity is positive, 27 m/s at an initial height of 35, so our equation becomes:
x = vt - gt^2 + 35
we want to know the time at which x=5, 30 meters below the cliff and 5 meters above the water
5 = 27t - 9.8t^2 + 35
solving for 't' we have t=3.6
approximately 3.6 seconds after being thrown, the rock will be 5 meters above the water