Flight time of a ball and maximum height ?

A ball is thrown at 25m/s

Find:

a) The maximum height
b) flight time

(There's nothing more so I assume the initial height is 0 and the angle 90)

ElectroNewby,

For constant acceleration "g" = 9.8 m/sec²

dV/dt = g

Integrating gives V = gt+Vo, where Vo is the initial velocity. Since gravity points down, the up direction is negative. Plugging in the values,

0 = gt -25 ==> t = 25/9.8 = 2.55 sec to go up and the same to go down

ds = Vdt = Vo*dt + gt*dt ===> s = Vo*t + gt²/2 + s1, where s1 = 0

Plugging in the values we get s = -25*2.55 + 9.8*2.55²/2 = -31.89 meters high

Ratch

Thank you very much Rachit !
That 25/9.8 was my missing puzzle and I should hit myself because it's simple

ElectroNewby said:

A ball is thrown at 25m/s

Find:

a) The maximum height
b) flight time

(There's nothing more so I assume the initial height is 0 and the angle 90)

Click to expand...

Hi,

Are you sure they meant 90 degrees because that's straight up. It could be, but most of the time these problems start with some known angle LESS than 90 degrees like 45 degrees so the projectile sweeps out an arc.