Honors 11th grade Physics 1-D Kinematics problem.?

Initial vertical velocity = 56 * sin 25 Initial horizontal velocity 56 * cos 25 Since the ground is level, the time for the spider to rise from the ground to the maximum height is equal to the time for the spider to fall from the maximum height to the ground. As the spider rises from the ground to the maximum height, its vertical velocity decreases from 56 * sin 25 m/s to 0 m/s at the rate of 9.81 m/s each second. Use the following equation to determine the time.

Vf = vi – a * t 0 = 56 * sin 25 – 9.81 * t t = 56 * sin 25 ÷ 9.81 This is approximately 2.42 seconds. Total time = 2 * 56 * sin 25 ÷ 9.8 = 112 * sin 25 ÷ 9.81 Use the following equation to determine the range. Range = v * cos?

* Total time Range = 56 * cos 25 * 112 * sin 25 ÷ 9.81= 6272 * cos 25 * sin25 ÷ 9.81 This is approximately 244.9 meters. Total time = 2 * v * sin ÷ g Horizontal distance = v * cos? * 2 * v * sin ÷ g = v^2/g * 2 * cos?

* sin? In trigonometry, 2 * cos? * sin?

= sin 2? Horizontal distance = v^2/g * sin 2? This is how the range equation is derived.

Horizontal distance = 56^2/9.81 * sin 50 This is approximately 244.9 meters. I hope this is helpful. The easiest way to determine the maximum height is to use the following equation.

D = ½ * (vi + vf) * t, vi = v * sin? , vf = 0, t = vi * sin? ÷ g d = ½ * v * sin?

* v * sin? ÷ g = (v * sin?)^2 ÷ (2 * g) Maximum height = (56 * sin 25)^2 ÷ 19.62 This is approximately 28.5 meters.

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