A circus performer juggling while standing on 10 a tightrope suspended 15.0 m above the ground tosses a sword directly upward into the air with a speed of 5.0 m/s. If the sword leaves his hand 1.0 m above the tightrope, what is the sword's maximum height above the ground? Round your answer to the nearest tenth

Respuesta :

The sword's maximum height above the ground rounded to the nearest tenth is 122.49m

In order to get the maximum height of the rope above the ground, we will use the formula for calculating the range;

[tex]R=u\sqrt{\frac{2H}{g} }[/tex]

Given the following parameters:

speed u = 5.0m/s

H is the maximum height

g is the acceleration due to gravity = 9.8m/s

R is the range = 15.0m

Substitute the given parameters into the formula to have:

[tex]15=5\sqrt{\frac{2H}{9.8} }\\\frac{15}{3}=\sqrt{0.2041H} \\5^2 = 0.2041H\\H = \frac{25}{0.2041}\\H= 122.48m[/tex]

Hence the sword's maximum height above the ground rounded to the nearest tenth is 122.49m

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