Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
A 0. 22 kg clay target is fired at an angle of 35° to the horizontal. It reaches a maximum height of 4. 6 m. Its initial speed when it is launched will be u = 16.66 m/s
Projectile motion is the motion of an object thrown (projected) into the air. After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory.
The object's maximum height is the highest vertical position along its trajectory. The horizontal displacement of the projectile is called the range of the projectile. The range of the projectile depends on the object's initial velocity.
h (max) = [tex]u^{2}[/tex] [tex]sin^{2}[/tex](theta) / 2g
given
theta = 35°
Maximum height = 4. 6 m.
u = ?
4.6 = [tex]u^{2}[/tex] * [tex]sin^{2}[/tex](35) / 2* 9.8
u = 16.66 m/s
To learn more about projectile motion here
https://brainly.com/question/11049671
#SPJ4
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.