A ball is thrown upward with a velocity of 30.0 m/s. Compute for: a. the time to reach the maximum height  b. the maximum height reached c. the time before the ball returns to the origin  d. the velocity with which the ball returns to the origin

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Let's assume that the initial height is zero. If not, nothing changes but the question b.: what is the maximum height reached over the initial height?

Denote the initial velocity as `V_0.` It is positive and upward. The velocity of the ball as a function of time is

`V(t)=V_0-g*t,`

and the height of the ball is

`H(t)=V_0t-(g*t^2)/2,`

where `g approx 10m/s^2` is the gravity acceleration.

 

The maximum height is reached when `V(t_1)` is zero. It is going to be negative after that moment, so the ball starts to fall.

`V(t_1)=V_0-g*t_1=0`

at `t_1=V_0/(g)=3(s).` This is the answer to the question a.

b. The maximum height is `H(t_1)=V_0*V_0/(g)-(g*(V_0)^2/(g^2))/2=(V_0^2)/(2g).`

In this case it is 900/20=45 (m).

 

c. The time `t_2` when the ball returns to the origin is when `H(t_2)=0,`  i.e.

`V_0t_2=(g*t_2^2)/2.`

The root `t_2=0` corresponds to the initial moment while `t_2=(2V_0)/(g)` =6(s) is the return time. It is no surprise that `t_2=2t_1` because the path of the ball is symmetrical.

 

d. `V(t_2)=V_0-g*t_2=V_0-g*(2V_0)/(g)=-V_0` =-30(m/s).
This means 30m/s downwards, also no surprise.

 

Of course all above is true if we ignore the air resistance (which is not entirely correct).

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