A long jumper leaves the ground at an angle of 20.0° above the
1 answer:
The main characteristics of projectile motion are
maximum height H =V²o sin² teta / 2g
the horizontal range R = V²o <span>sin² 2(teta) / g
so
1) the </span><span>horizontal direction he should jump
</span>R = V²o <span>sin² 2(teta) / g = 121 x 0.41 /9.8 = 5.06 m
</span>
2)<span>the maximum height reached
</span>H =V²o <span>sin² teta / 2g = 121*0.11 / 2 x 9.8=0.67m
</span>
3) for teta = 25°, R = V²o sin² 2(teta) / g= 7.24m, H= V²o <span>sin² teta / 2g=</span><span>1.1m
same method for the others
</span><span>graph , look at the image, and do the others like the example
</span>
maximum height H =V²o sin² teta / 2g
the horizontal range R = V²o <span>sin² 2(teta) / g
so
1) the </span><span>horizontal direction he should jump
</span>R = V²o <span>sin² 2(teta) / g = 121 x 0.41 /9.8 = 5.06 m
</span>
2)<span>the maximum height reached
</span>H =V²o <span>sin² teta / 2g = 121*0.11 / 2 x 9.8=0.67m
</span>
3) for teta = 25°, R = V²o sin² 2(teta) / g= 7.24m, H= V²o <span>sin² teta / 2g=</span><span>1.1m
same method for the others
</span><span>graph , look at the image, and do the others like the example
</span>
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