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A long jumper leaves the ground at an angle of 20.0° above the
horizontal and at a speed of 11.0 m/s.

1. How far does he jump in the horizontal direction?
2. What is the maximum height reached?
3. How far does he jump in the horizontal direction if the jumper leaves the
ground at angle 25°, 26°, 27°, 28°, and 29°?
4. What is the maximum height reached f the jumper leaves the ground at angle
25°, 26°, 27°, 28°, and 29°?
5. Plot several graphs for the following information:
a. Distance vs angle (Question 3)
b. Height vs angle (Que

Respuesta :

Nirina7
the main characteristics of projectile motion are

maximum height   H =V²o sin² teta  / 2g
the horizontal range  R = V²o sin² 2(teta) / g
so 
1) the horizontal direction he should jump

R = V²o sin² 2(teta) / g  = 121 x 0.41 /9.8 = 5.06 m

2)the maximum height reached
H =V²o sin² teta  / 2g = 121*0.11  / 2 x 9.8=0.67m

3) for teta = 25°,  R = V²o sin² 2(teta) / g= 7.24m, H= V²o sin² teta  / 2g=1.1m
   same method for the others
graph , look at the image, and do the others like the example
   
Ver imagen Nirina7

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