# Thread: A ball is thrown

1. ## A ball is thrown

A ball is thrown stright up with the speed 25m/s

how high does the ball go and at what time is it at half its height?

I think am suppose to use somthing like s=v*t but i dont know how to find the time or the s so i can solve for the other. I also know gravity should play a part in the speed

2. Originally Posted by JBswe
A ball is thrown stright up with the speed 25m/s

how high does the ball go and at what time is it at half its height?

I think am suppose to use somthing like s=v*t but i dont know how to find the time or the s so i can solve for the other. I also know gravity should play a part in the speed
i believe you are referring to suvat equations, those can work, but i never remember them. let's start from scratch

the acceleration due to gravity is -9.81 m/s

thus we have:

$\displaystyle a(t) = -9.81$ with the initial condition, $\displaystyle v(0) = 25$ m/s and $\displaystyle s(0) = 0$

now, let's find $\displaystyle v(t)$ and $\displaystyle s(t)$

(by the way $\displaystyle s(t)$ is the position function, $\displaystyle v(t)$ is the velocity function and $\displaystyle a(t)$ is the acceleration function)

so, $\displaystyle v(t) = \int a(t)~dt = \int -9.81~dt = -9.81t + C$

since $\displaystyle v(0) = 25$, we have:

$\displaystyle v(0) = 25 = -9.81(0) + C \implies C = 25$

thus, $\displaystyle v(t) = -9.81t + 25$

we obtain the max height when $\displaystyle v(t) = 0$

that is, when $\displaystyle t \approx 2.548$

plug that into the position function:

$\displaystyle s(t) = \int v(t)~dt = \int (-9.81t + 25)~dt = -4.9t^2 + C$

since $\displaystyle s(0) = 0$, we have that $\displaystyle C = 0$ here.

since we obtain the max height at $\displaystyle t = 2.548$, the max height is, $\displaystyle s(2.548)$

i think you can handle the rest

...by the way? have you done calculus? if not, the suvat equations are the way to go, forget everything i said and see here (note that these equations assume all initials are zero.