# Simple Laws of motion question

1. Oct 20, 2006

### faamfwvame

A friend and I were conducting a science experiment, where an object weighing roughly 550 grams was launched vertically, and after examining the video footage, was airborne for aproximatly 4.5 secconds. Having never studied physics before, I was curious if you could calculate the launch speed and or peak height from this information.

Any help would be highly appreciated.

Last edited: Oct 20, 2006
2. Oct 20, 2006

### SGT

Assuming negligible air resistence, wich is approximatelytrue if the object is a ball made of a dense material, the velocity of a body moving against gravity is $$v = v_0 - gt$$
At the top of the trajectory the velocity is 0 and the time is half the total time.

3. Oct 20, 2006

### faamfwvame

I don't really understand that sorry, but according to Wikipedia, a stationary object's speed increases by 9.81 metres per second as it falls. and since my object was falling for 4.5/2, 2.25 secconds, I figured the peak hieght would be 9.81 + (9.81x2) + (9.81x3)/4 which equals 36.7875 metres, which, having seen the object, sounds about right. And 9.81 being approximatly 22 miles per hour, I guessed the speed it was traveling when it hit the ground and indeed the speed it was launched at would be 22 x 2.25 which is 49.5 miles per hour.
However, having no expirience with physics, I am hesitant to trust these results. Could anyone confirm or correct this?

4. Oct 20, 2006

### SGT

I don't understand the calculations you made. The initial and final velocities are equal in absolute values.
$$v_0 = 9.81*2.25 = 22m/s = 11.4 mph$$
The maximum heogth will be
$$h = v_0*t - \frac{1}{2]g*t^2 = 24.8m$$