physics question

Question

An athlete executing a long jump leaves the ground at a 28.0 degree angle and travels 7.80m. (a) What was the takeoff speed? (b) If this speed were increased by just 5%, how much longer would the jump be?

Answer 1

Call his takeoff speed v.  This is a vector that points at an angle of 28° above horizontal. 

 

Start by resolving this vector into vertical and horizontal components.  The horizontal component (v_x) is v cos 28° and his vertical component (v_y) is v sin 28°.

 

Let's determine how long he was in the air.  Call that t.  Since his horizontal speed is not affected by gravity (we have to assume no air resistance), the distance of his jump is v_xt= vt cos 28°.  We know that's equal to 7.80 m, so we have an equation:  vt cos 28° = 7.80.

 

But what do we know about t?  Gravity exerts a force on him that slows down his vertical speed until he reaches his peak height, and then speeds him up again until he hits the ground.  And the time he spends rising is equal to the time he spends falling, so he rises for t/2 seconds.  During this time, gravity reduces his vertical speed from v_y to 0 at a constant rate during these t/2 seconds.  So gt/2 = v_y = v sin 28°.

 

Now we have two equations for t and v:

 

vt cos 28° = 7.80

gt/2 = v sin 28°

 

We can solve these. Start with the second equation and solve for t:

gt/2 = v sin 28°

t = 2v sin 28° / g

 

Substitute the above expression for t in the first equation.

v(2v sin 28° / g) cos 28° = 7.80

2v² / g sin 28° cos 28° = 7.80

(v² / g) 2 sin 28° cos 28° = 7.80

 

Notice that 2 sin 28° cos 28° is actually equal to sin (2*28°) = sin 56°.  So we have:

 

(v² / g) 2 sin 28° cos 28° = 7.80

(v² / g) sin 56° = 7.80

v² / g = 7.80 / sin 56°

v² = 7.80 g / sin 56° = 7.80 * 9.8 / 0.829 = 92.21

v = 9.60 m/s

 

So his takeoff speed was 9.6 m/s

 

If you add 5% to that, it would be 10.08 m/s.

 

Now we have to work back in the other direction.

 

His vertical speed would be 10.08 sin 28° = 4.73 m/s

 

Therefore his time to rise to his peak height would be 4.73 / g, and his total time in the air would be double that, or 9.46 /g = 0.965 seconds.

 

During that time, his horizontal speed would be 10.08 cos 28° m/s, so he would travel 0.965 * 10.08 cos 28° = 8.59 meters.

 

So the jump would be 8.59 - 7.80 = .79 meters longer.