Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

you take a running leap, leaving the ground with a speed 6.00 m/s at an anglle 40.0 degrees above horizontal.

Sagot :

The maximum height reached while running is calculated to be 0.75m.

The formula for the maximum height of the projectile motion is given by

H = u² (sin θ)²/2g

where, θ is the angle made with the horizontal

H is the maximum height reached

g is the acceleration due to gravity

u is the initial velocity

Given that, u = 6 m/s

g = 10 m/s²

θ = 40°

Substituting the above values, we have,

H = u² (sin θ)²/2g = 6² * (sin 40°)² / 2 * 10 = 36* 0.414/20 = 0.75 m

Thus, the maximum height reached while running is 0.75m.

To know more about height:

https://brainly.com/question/15416364

#SPJ4