Physics question?

Question

you jump from a diving board with a horizontal velocity of 3.5 m/s. If the diving platform is 6.0 above the water at what horizontal distance from the platform will you hit the water.

Answer 1

Basics: Y=.5*a*t^2, so t=sqrt(2*Y/a). Distance=3.5*t.
Answer, if you're truly lazy, is t=1.107 s, distance=3.873 m.
(I could be lying. Better check the math.)
Brian, you used the wrong starting formula for t; 2Y/a = t^2, not sqrt(t).

Answer 2

Remember, time in the air is COMPLETELY dependent on the Y direction. So, since the jump is described as horizontal, the initial Y velocity is 0. So, lets find out how long the jumper will be in the air.

Y = Vot + 1/2 at^2, Vo is initial Y velocity = 0, and the distance traveled in the Y direction, Y, is 6.0 m, and of course a = g = 9.8

Solving for t:

2Y/a = sqrt t and t = (2Y/a)^2
t=(2(6.0)/9.8)^2 = 1.50 seconds

Then using the time in the air and the X velocity, we can calculate the distance jumped:

X = Vot + 1/2 at^2

There is no acceleration, and we know the initial X velocity is 3.5
So, plugging in the values:

X = 3.5(1.5) + 1/2 (0)(1.5) = 5.25 m

Answer 3

The key point here is that the horizontal and vertical velocities and accelerations are completely independent!

Step 1. Use the 3rd equation of motion to find the *TIME* it takes for the diver to fall 6 meters.

s = ut + 1/2 at^2

s = the distance travelled from the initial state to the final state (displacement)
u = the initial speed
v = the final speed
a = the constant acceleration
t = the time taken to move from the initial state to the final state

rearranging this

t = sqrt(2s/a)
t = sqrt(2*6/9.8)

Where;
s = 6m (from diving board to water)
a = 9.8m/s^2 (the acceleration due to gravity)

Step 2. Using the 3th equation of motion calculate the horizontal distance the diver will travel before s/he hits the water.

s = ut + 1/2 at^2
s = 3.5 x sqrt(2*6/9.8) + 0
s = 3.87m

Answer 4

use 9.8m/s squared for the gravitational pull of earth