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A powerful grasshopper launches itself at an angle of 45° above the horizontal and rises to a maximum height of 1.01 m during the leap. With what speed v did it leave the ground? Neglect air resistance. Hint: Use conservation of energy. A. 6.29 m/s B. 7.15 m/s C. 5.98 m/s D. 6.72 m/s E. 5.37 m/s

Respuesta :

Answer:

6.29 m/s option (A)

Explanation:

theta = 45 degree, H = 1.01 m

let v be the launch speed

Use the formula for the maximum height for the projectile

H = v^2 Sin^θ / 2g

1.01 = v^2 x Sin^2(45) / (2 x 9.8)

1.01 = 0.0255 v^2

v^2 = 39.59

v = 6.29 m/s

onyebuchinnaji

The initial velocity of the grasshopper is 6.29 m/s.

Initial velocity of the grasshopper

The Initial velocity of the grasshopper is calculated from the following kinematic equation.

[tex]H = \frac{v_0^2 sin^2 \theta}{2g}[/tex]

where;

  • H is the maximum height
  • v is the initial velocity

[tex]v_0^2 = \frac{2gH}{sin^2\theta} \\\\v_0^2 = \frac{2 \times 9.8 \times 1.01 }{(sin45)^2} \\\\v_0^2 = 39.6\\\\v_0 = 6.29 \ m/s[/tex]

Thus, the initial velocity of the grasshopper is 6.29 m/s.

Learn more about initial velocity of a projectile here: https://brainly.com/question/12870645

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