Solving the Time for a Ball Thrown Vertically Upward

[Thiendrah]

Homework Statement

A ball is thrown vertically upward with an initial velocity at the edge of the roof of a building which is 30m above the ground. The speed of the ball is 14.5 m/s when the ball is 12m above the ground. How long does the ball take to reach maximum height?

Homework Equations

Vf^2=Vo^2+2ad

The Attempt at a Solution

First, i use that equation to find distance between the highest point and 12 above ground. Then I plan to use it one more time to find the initial velocity. But the thing is that

I find d=-10.727m as the distance between highest point and 12 above ground. This doesn't make sense, shouldn't it be higher than 18m?

 

[tiny-tim]

I find d=-10.727m as the distance between highest point and 12 above ground. This doesn't make sense, shouldn't it be higher than 18m?

Hi Thiendrah!

Show us your calculations, and then we'll know how to help.

 

[minifhncc]

v²=u²+2aS

Use that equation for all the calculation 12m above the ground and then add 12m to your final result.

So obviously v=0 (as it reaches the top) u=14.5ms^-1 a= -9.8ms^-1 (If up is positive), and then solve for S and add 12m.

 

[Thiendrah]

minifhncc, that's the same thing. You still get the same result.


0^2=14.5^2+2.(-9.8).S

S=10.727

10.727 + 12 = 22.727...

this doesn't make sense tho. The object is thrown vertically upward off a roof that's 30meters above the ground.

Am i doing anything wrong?

 

[tiny-tim]

minifhncc, that's the same thing. You still get the same result.


0^2=14.5^2+2.(-9.8).S

S=10.727

10.727 + 12 = 22.727...

this doesn't make sense tho. The object is thrown vertically upward off a roof that's 30meters above the ground.

Am i doing anything wrong?

Hi Thiendrah!

No … you seem to be right.

I can't see any way out of this in the question …

the question specifically says that the initial velocity is vertical …

your formula is correct (you could also have used KE + PE = constant, with the same result) …

the question is wrong.