Solving Projectile Motion Problem: 4.2 m/s

[fleur]

Homework Statement

Jim stands with a ball at the rear of a diving board, 10 m above the
surface of a swimming pool. His friend Bob is swimming below, at a
horizontal distance of 6 m from the front of the board. Jim thinks it
would be funny to roll his ball along the diving board so that it drops on
Bob’s head. Assuming that the effects of friction and air resistance are
negligible, calculate the speed (in m s-1) at which Jim must roll the
ball.

Homework Equations

s=u+1/2at^2
speed=distance/time

The Attempt at a Solution

rearranged s=u+1/2at^2 to find t
t=1.1s
speed 6/1.1=5.45m/s
However the answer is 4.20m/s
where did I go wrong?

 

[SteamKing]

Homework Statement

Jim stands with a ball at the rear of a diving board, 10 m above the
surface of a swimming pool. His friend Bob is swimming below, at a
horizontal distance of 6 m from the front of the board. Jim thinks it
would be funny to roll his ball along the diving board so that it drops on
Bob’s head. Assuming that the effects of friction and air resistance are
negligible, calculate the speed (in m s-1) at which Jim must roll the
ball.

Homework Equations

s=u+1/2at^2
speed=distance/time

The Attempt at a Solution

rearranged s=u+1/2at^2 to find t
t=1.1s
speed 6/1.1=5.45m/s
However the answer is 4.20m/s
where did I go wrong?

Why don't you show us your calculations of how you obtained t = 1.1 s for the time it takes the ball to fall from the board and hit the water.

 

[fleur]

Why don't you show us your calculations of how you obtained t = 1.1 s for the time it takes the ball to fall from the board and hit the water.

I did square root of 2x6/9.81

 

[SteamKing]

I did square root of 2x6/9.81

So the ball is falling horizontally? I thought the board was 10 m above the surface of the water.

 

[Clever Penguin]

The equation is wrong

Homework Equations

s=u+1/2at^2

The equation is s=ut+1/2at²

I did square root of 2x6/9.81

The time taken for it to hit the pool when dropped is the same as when rolled since the horizontal and vertical components of the velocity do not affect each other.