Question

a diver on a board 7.5m above the water walks off the end with a horizontal velocity of 2.3 m/s when they hit the surface of the water how far away from the edge of the board will they land ?

Answer 1

Answer:

2.846m

Explanation:

The diver is performing projectile motion.

To find x(final), we are going to use the equation x(final) = v(initial)*t + x(initial)

x(initial) = 0

x(final) = ?

v(initial) = 2.3 m/s

we don't know t

To find t we will use y(final) = 1/2*(-9.8)*t^2 + v(initial in the y dir.)*t + y(initial)

- 9.8 in the acceleration in the y dir.

y(final) = 0

y(initial) = 7.5

v(initial in the y dir.) = 0

If we solve for t we get: t = 1.237s

Now we have all the components to solve for x(final) in x(final) = v(initial)*t + x(initial)

x(final) = 2.3*1.237 + 0

x(final) = 2.846m