Projectile Motion with Given Angle and Velocity

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The discussion revolves around solving a projectile motion problem involving a bullet shot at a 27.2-degree angle with an initial velocity of 412 m/s from a height of 2 meters. The user has successfully calculated the bullet's time in the air as 38.4 seconds and its range as 14,079.67 meters but is struggling with finding the maximum height and the time to reach that height. The solution for the time to maximum height can be derived using the equation t = Vo * Sin(θ) / g. To find the maximum height, the equation Y = 2 + ½gt² can be applied. The discussion highlights the importance of using the correct equations to solve projectile motion problems effectively.
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I am having a problem with this question. I have it half solved but I cannot figure out what equation I am supposed to use for the rest.

Homework Statement


A hunter shoots his rifle at an angle of 27.2 degree above the horizontal. He is holding the rifle so that the muzzle of the rifle is exactly 2 meters above the ground. The muzzle velocity of the rifle is 412 m/s

The answers needed for this question are the following:
How long is the bullet in the air?
What is the bullet's range?
What is the maximum height of the projectile?
How long does it take the bullet to reach its maximum height?

Starting velocity = 412 m/s
Starting height = 2 meters
Range = 14079.67
change in time = 38.4 seconds

Homework Equations



y=y0+Y0y\Deltat+1/2(ay)\Deltat2
range=v0(cos27.2)\Deltat

The Attempt at a Solution


I have found the answers to the first two answers which are:
How long is the bullet in the air? 38.4 seconds
What is the bullet's range? 14079.67 meters

but I am not sure what equations I am supposed to use for these questions:
What is the maximum height of the projectile?
How long does it take the bullet to reach its maximum height?
 
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Welcome to PF.

You can calculate the time to max height easily enough.

Vo*Sinθ = g*t

t = Vo*Sinθ/g

You should know then that Y = 2 + ½gt²
 
thank you
 
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