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An object is launched at 13.8 m/s from the ground. The equation for the object's height "h" at time "t" seconds after launch is: h(t) = -4.9t² + 13.8t, where h is in meters.
theredeemer said:An object is launched at 13.8 m/s from the ground. The equation for the object's height "h" at time "t" seconds after launch is: h(t) = -4.9t² + 13.8t, where h is in meters.
The formula for calculating the height of an object launched at 13.8 m/s is h = (v^{2}sin^{2}θ)/2g, where h is the height, v is the initial velocity (13.8 m/s), θ is the angle of launch, and g is the acceleration due to gravity (9.8 m/s^{2}).
The angle of launch can be determined by using a protractor to measure the angle at which the object is launched from the ground. It is important to measure the angle accurately, as even a small difference can greatly affect the calculated height.
No, the height of the object is also affected by the angle of launch and the acceleration due to gravity. A higher initial velocity and a smaller angle of launch will result in a higher height, while a lower initial velocity and a larger angle of launch will result in a lower height. The acceleration due to gravity remains constant.
Yes, the formula can be used for any object launched at 13.8 m/s, as long as the angle of launch and the acceleration due to gravity are known. However, it is important to note that the formula is derived for objects launched from a horizontal surface and may not be accurate for objects launched from other surfaces.
This formula can be applied in real-life situations such as calculating the height of a projectile launched from a catapult or a rocket launched into the air. It can also be used in sports such as basketball or baseball to determine the height of a ball thrown or hit at a certain velocity and angle. Additionally, this formula is used in engineering and physics to analyze the motion of objects.