A diver achieves a horizontal velocity of 3.75 m/s from a diving platform located 6.0 m above the water. How far from the edge of the platform will the diver be when he hits the water?
2006-12-03 15:05:42 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
A diver achieves a horizontal velocity of 3.75 m/s from a diving platform located 6.0 m above the water. How far from the edge of the platform will the diver be when he hits the water? What is the formula used? Thanks for your time!
2006-12-03 15:17:11 · update #1
Ok, we know that the car travels 6.0m down to hit the ground, let's see how long it would take him to fall that distance, using g=9.8m/s^2 the acceleration of gravity.
Y=(1/2)gt^2 , rearrange to get (2Y/g)^(1/2) = t
since y=6 this give t=(12/9.8)^(1/2) ot t= 1.1066s
It takes 1.1066s to fall from a height of 6m.
Keep in mind that vertcal and horizontal components do not mix with each other! At 1.1066s he hits the ground no matter how fast he is travelling in the horizontal direction!
The horizontal distance then is X=vt , where v=3.75 and t is what we found t=1.1066
X=(3.75)(1.1066)=4.1498m m the distance he travelled from the edge.
Hope this helps!
2006-12-03 15:25:37 · answer #1 · answered by William M 2 · 0⤊ 0⤋
s = s0 + v0t + (1/2)at^2
t = √2s/a
t = √(2*6/9.80662)
d = ut
d = 3.75√(2*6/9.80662)
d = 4.15 m
2006-12-03 15:18:39 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
1.6 m
2006-12-03 15:15:28 · answer #3 · answered by Aditya N 2 · 0⤊ 1⤋