| Dementia
loves to dive from the high cliffs by the sea. When she dives, she runs
straight forward with a velocity of 10 m/s right off the edge of the cliff.
She has to be sure she runs with enough speed, however, because there are
sharp, jagged rocks below.
If the cliff is 40 m above the water and we disregard air resistance (and the owl stays out of the way), how far will Dementia travel (horizontally) from the base of the cliff? |
We know Dementia's horizontal velocity,
so we need to find the time it takes her to fall 40 m. From
dy = 0.5ayt2, we get t = (2dy/g)1/2 where g is the acceleration due to gravity. t = [(2)(40m)/(9.8m/s2)]1/2 = 2.9 s dx = vx t = (10m/s)(2.9s) = 29 m |