Kinematic formula for projectiles

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To determine the maximum height of a projectile launched at 20.0 m/s at a 27.0-degree angle, the initial vertical velocity is calculated as 9.08 m/s. The time of flight is essential for solving the problem, which can be found using the vertical motion equations. The horizontal distance can be calculated using the horizontal velocity of 17.8 m/s and the time of flight. For other projectile questions, the designation of final velocity depends on the context, such as whether the object is at its peak or returning to the ground. Understanding these principles is crucial for solving projectile motion problems effectively.
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Q: an object is thrown from the ground into the air with a velocity of 20.0m/s at an angle of 27.0 degrees to the horizontal. what is the max height reached by this object?

i drew the triangle and found vertical side to be 9.08 m/s and horizontal side to be 17.8m/s using sin and cos.
/|
20 / | 9.08
/__|
17.8

but the part I'm having trouble with is how to do the horizontal part of the equation using d=vt. I am assuming that i have to use that eqn to find time so i can use that value for the vertical part in finding d, using kinematic formula.
in the horizontal part d=vt, what would i use as d? i have v as 17.8

also a side note: for other projectile questions, when would you designate final velocity as unknown, zero, or the negative sign of the initial velocity (ie. initial velocity=14, final velocity = -14) ?
 
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The key to this problem is to find the time of flight by knowing the initial vertical velocity. From there you'll be able to find time of flight, horizontal distance, vertical distance.
 
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