Posted by **Zeek ** on Sunday, February 5, 2012 at 12:53am.

An athlete executing a long jump leaves the ground at a 32.5° angle and travels 7.72 m. What was the take-off speed?

If this speed were increased by just 4.9 %, how much longer would the jump be?

- Physics -
**drwls**, Sunday, February 5, 2012 at 3:57am
The formula you need to answer both questions is

Distance = Vo^2*sin(2A)/g

where A it the takeoff angle.

Deriving that formula would be a good exercise for you. You would use the trig identity sin(2A) = 2 sinA cosA.

In the first case,

Vo^2 = g*7.72m/sin65 = 83.56 m^2/s^2

Take the square root of that for Vo.

Increasing the speed by a factor 1.049 will increase the length of the jump by a factor (1.049)^2 = 1.100

(a 10.0% increase).

## Answer this Question

## Related Questions

- physics - An athlete executing a long jump leaves the ground at a 30.3 angle and...
- physics - An athlete executing a long jumper leaves the ground at a 27.0 degree ...
- physics - An athlete executing a long jump leaves the ground at a 32.3 angle and...
- physics - An athlete executing a long jump leaves the ground at a 28.9 ∘ ...
- physics - An athlete executing a long jump leaves the ground at a 20° angle and ...
- physics - An athlete executing a long jumper leaves the ground at a 27.0 degree ...
- math/ physics - An athlete executing a long jumper leaves the ground at a 27.0 ...
- science - An athlete executing a long jump leaves the ground at a 27.7∘ ...
- physics - An athlete executing a long jump leaves the ground at a 28.0 angle ...
- physics. HELP PLEASE ! - An athlete executing a long jump leaves the ground at a...

More Related Questions