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1. An object is launched at a velocity of 20 m/s in a direction making an angle of 25 degrees upward with the horizontal. a) What is the maximum height reached by the object? 3.57 m b) What is the total flight time (between launch and touching the ground) of the object? 1.69 s c) What is the horizontal range (maximum x above ground) of the object? 30.63 m d) What is the magnitude of the velocity of the object just before it hits the ground? 20 m/s

User Maykeye
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Answer:

a) To find the maximum height, you can use the formula for vertical motion: \( \text{max height} = \frac{v^2 \sin^2(\theta)}{2g} \), where \( v \) is the initial velocity, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity. Substituting the given values, you get \( \text{max height} = \frac{20^2 \sin^2(25^\circ)}{2 \cdot 9.8} \). Calculate to find the result.

b) The total flight time can be found using the formula \( \text{total time} = \frac{2v \sin(\theta)}{g} \). Plug in the values to get \( \text{total time} = \frac{2 \cdot 20 \cdot \sin(25^\circ)}{9.8} \).

c) The horizontal range is given by \( \text{range} = \frac{v^2 \sin(2\theta)}{g} \). Plug in the values to get \( \text{range} = \frac{20^2 \sin(2 \cdot 25^\circ)}{9.8} \).

d) The magnitude of the velocity just before hitting the ground is \( v \cos(\theta) \). Plug in the values to get \( 20 \cos(25^\circ) \). Calculate to find the result.

Perform the calculations to get the answers for each part.

User Sandoz
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