Posted by **Cooper** on Thursday, September 27, 2007 at 7:44pm.

A ball is kicked toward Ryan with an initial velocity of 21.0 m.s at an angle of 41 degrees above the horizontal. At that instant, Ryan is 51.0m from the kicker. In what direction, and with what constant velocity, should Ryan run in order to catch the ball at the level it was kicked at?

- Twelfth Grade Physics -
**Cooper**, Thursday, September 27, 2007 at 8:07pm
Just confused a little bit as to where to start. Can the 50 m also be the Range of the ball? And if that is so, can you use that to find v2, because the ball is falling at a constant of 9.81 m/s^2...and then you could say that Ryan runs the same velocity as the ball but in the opposite direction...is this wrong? I need a little push in the right direction!

- Twelfth Grade Physics -
**bobpursley**, Thursday, September 27, 2007 at 8:47pm
Find out where the ball lands.

time in air:

hfinal=viy*t -1/2 g t^2

where vi=21*sin41, and hfinal=0

solve for time in air.

Then, distance:

x= vix*timeinair where vix=21cos41

Now you know where the ball will land.

Ryan has to start at 51m point, which direction does he run?

## Answer This Question

## Related Questions

- Physics - 2-D kinematics - A soccer ball is kicked with an initial horizontal ...
- physics - A ball of mass 0.4 kg, initially at rest, is kicked directly toward a ...
- physics - A soccer ball is kicked with an initial horizontal velocity of 16 m/s ...
- Physics - 4. A ball is kicked toward a fence from a point 32 meters away. The ...
- Physics - A soccer ball is kicked with an initial horizonal velocity of 18 m/s ...
- physics - A soccer ball is kicked into the air at an angle of 38 degrees above ...
- Physics - A ball of mass 0.6 kg, initially at rest, is kicked directly toward a...
- physics - a soccer ball is kicked into the air at an angle of 38 degrees above ...
- physics - a soccer ball is kicked into the air at an angle of 38 degrees above ...
- physics - a soccer ball is kicked into the air at an angle of 38 degrees above ...

More Related Questions