Physics Projectiles Help?
Cliﬀ divers at Acapulco jump into the sea from a cliﬀ 36.4 m high. At the level of the sea, a rock sticks out a horizontal distance of 13.16 m. The acceleration of gravity is 9.8 m/s2. With what minimum horizontal velocity must the cliﬀ divers leave the top of the cliﬀ if they are to miss the rock?
Answer in units of m/s.
- 1 decade agoFavorite Answer
vf^2 = 2gd -> d = 36.4 m -> vf = sqrt(2*9.8*36.4) = 26.71 m/s
vf = 1/2 * g * t -> t (the time it takes from start to hit the water) = 2*vf/g = 2 * 26.71/9.8 = 5.45s
If the dude falls straight down, he'll hit the water in 5.45s, so that's how long he will be in the air so long as he is not given any initial upward velocity. In this case, the initial velocity is horizontal and the horizontal velocity will remain constant throughout the jump (assuming no air resistance)
So he needs to clear 13.16m in 5.54 s or 13.16/5.54 m/s = 2.414 m/s
Therefore, so long as his initial horizontal velocity is greater than 2.414 m/s, he'll miss the rock
I hope... :)Source(s): My defective brain
- JimLv 71 decade ago
Time to fall 36.4 m ignores air drag = t =√2d/g =√(2)(36.4)/9.81 =√7.42099898 = 2.72s
To travel 13.16 m in 2.72 s takes 13.16/2.72 = 4.84 m/s horizontal velocity ANS