An athlete executing a long jump leaves the ground at a 20° angle and travels 7.30 m. What is takeoff speed?

Question

An athlete executing a long jump leaves the ground at a 20° angle and travels 7.30 m. What was the takeoff speed? If this speed were increased by just 3.0 percent, how much longer would the jump be?

An athlete executing a long jump leaves the ground at a 20° angle and travels 7.30 m. What is the takeoff speed (aka initial velocity)? If this speed were increased by just 3.0 percent, how much longer would the jump be?

Answer 1

0 - nada - stopped. In order to do a long jump, you have to start running from a dead stop sometime.
Don't like my smart alec answer? Well, it seems to be a trick question. The distance jumped does not entirely depend on speed.

Answer 2

Assume 5.80m is horizontal There is a range formula you can use but teachers often don't want you to use it, so I'll give it an end. Lets say time taken is t vx = 5.80/t delta y = 0 so 0 = viyt + 1/2 g t^2 or t = 2viy/9.8 so vx = 5.8*9.8/(2viy) or vxviy = 5.8*9.8/2 = vcos20vsin20 solve v^2 = 5.8*9.8/(2*cos20*sin20) for v The formula is Range = v^2sin(2theta)/g Increasing v by 5% means range increases by 10% So 5.80 + .580 To show it, add 5% to v found earlier and find new range

Answer 3

Let the takeoff velocity = v0, The vertical component is then v0y = v0*sinø and the horizontal v0x = v0*cosø. The total time of the jump is twice the time to top of trajectory (time up tu = time down). The top of trajectory is reached when the verical velocity = 0. The vertical velocity is v0y - g*tu, so tu = (v0*sinø)/g, and total jump time is t = 2*(v0*sinø)/g. In that time he has traveled L horizontally, which is his horizontal speed time total time, L = v0x*t = (v0*cosø)*t. Plugging in for t you get

L = v0*cosø * 2*(v0*sinø)/g = 2*v0^2*sinø*cosø/g

Solve for v0 = √[L*g/(2*sinø*cosø)]

Since L is proportional to v0^2, a 3% increase in v0 will result in a 6% increase in L.