# Two Dimensional Motion (I)

• Aggie

#### Aggie

I am stumped big time!

Two runners start at the same point on a straight track. The first runs with a constant acceleration so that he runs 93 yards in 8 seconds. The second runner waits 3 seconds and then throws a rock at his opponent's head. If the head and the rock are at the same level form the grownd, what must the inital magnitude of the velocity be if the rock is to hit the head just at the 93 yard tape?

What distance does the rock need to cover in how much time?

The same distance as the runner who actually runs covers but the time is unknown

Is it? The stone must hit the runner as he reaches the finish line. You know the time that this happens, so can you say what the time that the stones takes in the air is?

the stone takes 5 secs in the air

Yup, so now can you set up a relevant equation?

how do you get the vertical component of velcocity.

you get the horizontal component of velocity with what you have - 93/5 = 18.6 yrds/sec

For the horizontal component, you have used the equation d=ut+at2/2, where u is the initial velocity. Set up a similar equation in the y direction; you know d, a and t, so you can solve for uy.

the vertical height is not given. unless we use zero
then we get 24.5

Please can you explain this because it makes no sense. We don't know the vertical distance