How long does it take a ball to be back

[mimi.janson]

Homework Statement

A ball is trown vertically straight up in the air. After t seconds the ball has reached the higth

Question : after how many sconds is the ball back to the spot it was trown from ?

Homework Equations

I have figured out that i need to differentate.

The Attempt at a Solution

So i thought i had to solve it this was :

60t – 5t² --> 60-10t but it seems like this isn't the answer

besides i have made a drawing of the information i got . The orange line shows how the ball is thrown straight up, and the blue shows that i have to solve how much that and the orange are when they are added when reffereing to time in seconds .

 

[PeterO]

Homework Statement

A ball is trown vertically straight up in the air. After t seconds the ball has reached the higth s(t) = 60t – 5t² in metres

Question : after how many sconds is the ball back to the spot it was trown from ?

Homework Equations

I have figured out that i need to differentate.

The Attempt at a Solution

So i thought i had to solve it this was :

60t – 5t² --> 60-10t but it seems like this isn't the answer

besides i have made a drawing of the information i got . The orange line shows how the ball is thrown straight up, and the blue shows that i have to solve how much that and the orange are when they are added when reffereing to time in seconds .

one formula for vertical motion is S = ut + 1/2 gt² (acceleration under g)

By comparison to s(t) = 60t – 5t²

we can see the initial upward velocity is 60 m^-1 and g is taken as 10 ms^-2.

In that case you can easily establish how long it takes to stop moving up, and it takes that much again to come back down.

EDIT: you don't need differentiation - just find the values of t when S(t) = 0 - and one of those times is at time zero - when it all began.

 

[mimi.janson]

one formula for vertical motion is S = ut + 1/2 gt² (acceleration under g)

By comparison to s(t) = 60t – 5t²

we can see the initial upward velocity is 60 m^-1 and g is taken as 10 ms^-2.

In that case you can easily establish how long it takes to stop moving up, and it takes that much again to come back down.

EDIT: you don't need differentiation - just find the values of t when S(t) = 0 - and one of those times is at time zero - when it all began.

ok thank you