Question

PLEASE HELP

A juggler tosses a ball from her hand at a height of 1 meter with an initial upward velocity of 10 m/s.

After how many seconds will the ball reach its highest point?

I think the equation would be
h= 4.9 m^2 + 10 t + 1 (I think)

Answer 1

The equation you have written gives the HEIGHT at any time and is correct.

However - the question asks you about time.

The maximum height is gained when the velocity is 0 (i.e. it has stopped going up and is about to fall)

the equaion you need is

v = u + at

v=0
u = 10m/s
a= -9.8 m/s^2

so you can work out t

Answer 2

equation should be
h=- 4.9*t^2 m^2 + 10 t + 1 (I think)

Answer 3

in general
y = 1/2*at^2 + vt + y0
here a = -g, since acceleration due to gravity points downwards
y = 1/2 (-9.81)*t^2 + 10*t + 1
y = -4.9t^2 + 10*t + 1

Answer 4

as mr nood pointed out, you can use the equation
vf = vi + at
where vf = 0
vi = 10,
a = -9.8
0 = 10 + -9.8*t